• Overview
• Alternating Groups
• A Simple Group Contains Even Permutations
• Generated by p Cycles
• Simple Groups up to 200
• p Cycles and Normality
• Group of Order 168
• Special Linear, Center
• p Sylow Subgroups
• Eigen Values of 1, All or None
• Eigen Values in F
• Special Linear Group is Simple, n = 2
• Special Linear Group is Simple

A simple group has no normal subgroups other than itself and 1, somewhat like a prime number.

If p is prime, the cyclic group of order p is simple, because it has no subgroups at all. This is a "class" of simple groups, one for each prime p. There are other classes of simple groups, and some sporadic simple groups that stand alone.

Over 100 mathematicians have worked together to classify all the finite simple groups, publishing hundreds of articles comprising thousands of pages. In 1985, Scientific American described this accomplishment in an article entitled "An Enormous Theorem." Obviously I'm not going to present all that material here.

As mentioned above, cyclic groups of prime order are simple. Alternating groups and special linear groups are also simple. I will address these classes, and a couple more groups, and then call it a day.

The symmetric group on n letters, denoted S_n, consists of all permutations on n letters. The alternating group on n letters, denoted A_n, consists of all even permutations on n letters. A_n is normal in S_n, since it is a subgroup of index 2; and for n ≥ 5, A_n is simple.

A₁ and A₂ are trivial, and A₃ is the 3 cycle. Let's look at G = A₄.

By the Sylow theorems, G has a subgroup of order 4. Call this subgroup H. If G permutes the letters abcd, H consists of abcd, badc, cdab, and dcba. These are the double transpositions. (A 4 cycle is impossible, since that would be an odd permutation.) So H is isomorphic to Z/2 * Z/2.

The elements not in H are the eight 3 cycles. Since H is the only sylow subgroup it is normal in G, and the factor group is Z/3. G also contains Z/3, and 3 and 4 are relatively prime, so G is a semidirect product.

Let G = A_n for n ≥ 5, and show that all the permutations in G can be generated by the three cycles that permute 1 2 and k, for every k between 3 and n. Given k and l, pull k back to position 2, then push k out to position l (moving l into position 1), then rotate l into position k. This swaps 1 and 2, as it swaps k and l.

Given a random permutation, take these steps to bring it back to start. If 1 and 2 aren't in the first 2 slots, either in position or swapped, put them there in at most two steps. If 3 isn't in place, swap it with the element in position 3, so that 3 is where it belongs. This will swap the first two elements. Do this for 4 and 5 and so on through n. Since all permutations are even, 1 and 2 will be in position when you're done.

Note that the selection of 1 and 2 in the above was arbitrary. The 3 cycles tied to any two positions will generate G.

Suppose H is a normal subgroup of G. Let H contain q, the 3 cycle on abc shifting left, and let x be the involution that swaps a with b and c with k. Now xq/x leaves c where it is and cycles a b and k. This must appear in H, and similarly for every k, thus generating G. If H is a proper normal subgroup, it does not contain an isolated 3 cycle.

Suppose q is in H and q includes a cycle of length r, for r ≥ 4. Let x be the three cycle that shifts the first three members of the r cycle. Let x shift left and q shift right. By normality, (xq/x)/q belongs to H. The r cycle has been turned into a 3 cycle. Other cycles within q have been pushed forward and pulled back, and are no longer present. Thus H contains an isolated 3 cycle, and H is all of G. Henceforth all cycles in the permutations of H have length 2 or 3.

Next suppose H has q with at least two disjoint three cycles, and let x permute the first two members of one cycle and the first member of the other. Let x and q both shift left. Now (xq/x)/q is a 5 cycle in H, and by the above, H = G.

Next suppose q contains a three cycle and at least one two cycle. Since q² is a three cycle, H = G.

Next suppose q contains more than two two cycles. If x is a three cycle that permutes a pair and a member of another pair, then w = xq/x/q exchanges one pair for another, and leaves everything else alone. That leaves a double transposition.

Since n > 4, let z be a three cycle that permutes the entries in one of the pairs with an entry not in either pair. Now zw/z/w is a three cycle in H, and H = G.

A single transposition is impossible, as that is an odd permutation.

In each case H = G, so G is simple.

The same proof shows A_∞ is simple, where A_∞ is the finite even permutations on an infinite set of letters.

If S_n has some other normal subgroup H, intersect H with A_n, giving a group that is normal in S_n, and in A_n, which is impossible, since A_n is simple. Actually there is a catch; H might intersect A_n in the identity element. But that means x in H cannot be applied twice, for that would be an even permutation living in A_n, unless every element of H is an involution. If there are two involutions, combine them to get something in A_n, hence H is a single involution, which we will call t. If t is 3 or more transpositions, combine with a 3 cycle, as shown above, to get a double transposition, which lives in A_n. Thus t is a single transposition. Conjugate t by a 3 cycle, then apply t again to get a 3 cycle. Thus there are no other normal subgroups of S_n.

The same proof shows A_∞ is the only nontrivial normal subgroup of S_∞.

Let G be a permutation group with at least one element x that is an odd permutation. Let H = G∩A_n. Thus H is the elements of G that are even permutations. Multiply H by x to get a map from H onto the elements of G that are odd permutations. This is a bijection, hence there are just as many elements in H as there are outside of H. Since H has index 2 it is normal in G. Seen another way, H is the kernel of the parity homomorphism.

If G is simple then H is trivial. This means G is a single involution, an odd number of disjoint transpositions.

A simple group, beyond Z/2, contains only even permutations, in any representation of that group.

Let G be a permutation group in S_n. Let T, a subset of the n elements, be a block if G maps T onto T or G maps T to a set of elements disjoint from T. The block stays where it is, or it moves entirely somewhere else. Obviously any single element acts as a block. If a block has size 1 or n, it is trivial.

G is primitive if all blocks are trivial.

If G is not trivial and not transitive, G is not primitive either. Find something that moves and let its orbit define a block T. G always keeps T within T, and since G is not transitive, T is a proper block, whence G is not primitive.

The converse is not true. Let G be the symmetric group on one set, and on another set in parallel, and let G also swap the two sets. Each set is a block. G is transitive, but not primitive.

Assume G is transitive, and is generated by one or more isolated prime cycles. Suppose T is a nontrivial block, and let c be one of the generators. In other words, c is an isolated p cycle. We know that c, invoked again and again, eventually produces the identity permutation. Let k be the smallest exponent such that c^k brings some of the elements of T back into the block. If they don't all come back then c^k shows that T is not a block after all. So they are all back in the block. They may or may not be back in their original positions, but coming back to the block is necessary for coming back into position, so k divides p. In fact k = 1 or p. If k = p then they come back into place after p steps. Think of c as the entire group. Each point of T seeds an orbit of length p that comes back to that point, and does not touch T otherwise. Orbits are disjoint, thus c is several p cycles running in parallel, not an isolated p cycle. Yet c is a generator, one p cycle by assumption, thus k = 1, and c fixes T or moves the elements of T about.

Since G is generated by these cycles, G keeps T within T. Yet G is transitive, so T is all of S. A transitive group generated by p cycles is primitive.

If G includes a transposition, and G is primitive (hence transitive), G is all of S_n. Prove this by induction on k, as k runs from 2 to n.

Since there is a transposition, G includes the subgroup S₂. This is our starting point, k = 2.

Let G include S_k, permuting the k elements in a subset U of S. Let V be the complement of U in S. If a transposition swaps something in U for something in V, G contains S_k+1. If there is no such transposition, we have more work to do.

Let H be the subgroup of G generated by all transpositions. These transpositions stay within U, or within V. H maps U to U and V to V. Thus H has at least one nontrivial orbit in U, and perhaps more in V.

Take a moment to verify that the conjugate of a transposition is another transposition, whether that transposition is part of no cycles, adjacent in one cycle, nonadjacent in one cycle, intersecting a cycle, or shared between two cycles. Thus H is normal in G.

Select an orbit under H, contained in U, and call it T, since T is going to become a block. Let r, a permutation in G, move x to y, where x and y are in T. Where does r take z if z is also in T? Let q be the permutation in H that takes z to x. Thus qr takes z to y. Let r take z to w. Now r inverse times qr takes w to y, and since H is normal, this action is part of H. Thus w lies in T. Therefore r keeps z within T, and z was arbitrary, so r maps T onto T. If r moves any x to some y in T then r maps T onto T. Failing this, r moves T entirely somewhere else. This holds for all r in G, hence T is a block.

Since G is primitive, there can be no proper blocks. This is a contradiction, hence some transposition swaps elements between U and V, and we can take the next step from k to k+1. Ratchet k all the way up to n, and G = S_n.

Here is a variation on the above. Assume G is primitive and contains a 3 cycle. This means G contains A₃, i.e. k = 3. Proceed by induction on k as before.

A 3 cycle across U and V allows us to extend U by either 1 or two elements. Take a moment to verify this; it's just shuffling things about.

If there is no 3 cycle straddling U and V, let H be the subgroup generated by all the 3 cycles of G. The conjugate of a 3 cycle is a 3 cycle, whether that 3 cycle lives in 0, 1, 2, or 3 cycles within the conjugating permutation. I'll pause while you verify this; it's not a snap.

• intersects no cycles
• shares a with a cycle
• shares a with a cycle and b with another cycle
• shares a with a cycle and b with another cycle and c with a third cycle
• shares ab adjacent within a cycle
• shares ab adjacent with one cycle and c with another
• shares ab with a transposition
• shares ab with a transposition and c with another cycle
• shares a b nonadjacent in a cycle
• shares a b nonadjacent in a cycle and c in another cycle
• abc with the same 3 cycle or its inverse
• abc adjacent and in the same direction as a larger cycle
• abc adjacent and opposite in a larger cycle
• ab adjacent and c apart in a larger cycle
• a b and c apart in a larger cycle

Since H is normal in G, run the same orbit-block argument as above. You get the same contradiction, hence we can ratchet k all the way up to n, giving A_n.

The corollary is more important than the theorem. If a permutation group on p elements includes a p cycle and a 2 cycle, the group generated by these two prime cycles is primitive, and transitive, and therefore all of S_p.

Similarly, in a set of p elements, a p cycle and a 3 cycle generate A_p. If an odd permutation is tossed in, then the group becomes S_p.

If G acts on 5 elements and includes a 5 cycle and a double transposition, then all the permutations are even, and G could once again be A₅, but it could also be D₅. Arrange the letters ABCDE so they shift left and right, and assume A does not take part in the transpositions. B could swap with C D or E. If B swaps with E, and C swaps with D, then we are reflecting a pentagon through a line drawn from A to the midpoint of CD. This produces the dihedral group. If B swaps with C and D swaps with E, then shift right and apply the swaps, and find a 3 cycle. If B swaps with D and C swaps with E, then shift right twice and apply the swaps, and find a 3 cycle. These arrangements produce A₅.

What are all the simple groups with order ≤ 200? The prime cyclic groups, and A₅ for sure. There aren't any others, except for 168.

Skip over p groups; they all have nontrivial centers, which are normal subgroups. This rules out orders like 25 and 27.

If the order of the group is np^k, and there is no r dividing n with r = 1 mod p, other than 1, then the sylow subgroup of order p^k is unique by the third Sylow theorem, and is a proper normal subgroup. This rules out orders like 20, 35, and 84.

If |G| = np^k, and |G| does not divide n!, then the group cannot be simple, as per the strong Cayley theorem. This rules out orders like 12, 36, and 80. Actually |G| must divide n!/2, since a simple group cannot contain odd permutations. Example order 112 = 7×16, which divides into 7!, but not 7!/2.

1. trivial

2. prime

3. prime

4. prime power

5. prime

6. third sylow 3

7. prime

8. prime power

9. prime power

10. third sylow 5

11. prime

12. strong cayley 4

13. prime

14. third sylow 7

15. third sylow 5

16. prime power

17. prime

18. third sylow 9

19. prime

20. third sylow 5

21. third sylow 7

22. third sylow 11

23. prime

24. strong cayley 8

25. prime power

26. third sylow 13

27. prime power

28. third sylow 7

29. prime

30. If Z/3 and Z/5 are not normal in the simple group of order 30, there are 24 elements of order 5 and 20 elements of order 3, exceeding 30.

31. prime

32. prime power

33. third sylow 11

34. third sylow 17

35. third sylow 7

36. strong cayley 9

37. prime

38. third sylow 19

39. third sylow 13

40. third sylow 5

41. prime

42. third sylow 7

43. prime

44. third sylow 11

45. third sylow 5

46. third sylow 23

47. prime

48. strong cayley 16

49. prime power

50. third sylow 25

51. third sylow 17

52. third sylow 13

53. prime

54. third sylow 27

55. third sylow 11

56. There are 8 subgroups of order 7, consuming 8×6 = 48 distinct elements of order 7. This leaves just enough for the subgroup of order 8, which is required. This sylow subgroup is unique, hence normal in G.

57. third sylow 19

58. third sylow 29

59. prime

60. If G is simple it has 6 sylow groups of order 5, 10 sylow groups of order 3, and 15 sylow groups of order 4. Yes, the sylow theorems permit 4 sylow groups of order 3, but that creates a normalizer of index 4, whence G embeds in A₄, which is impossible. Similarly, we could havfe 5 sylow groups of order 4, but that places G in A₅, whence G = A₅. So assume 10 groups of order 3 and 15 groups of order 4.

    If the groups of order 4 are disjoint, G has 24 elements of ordder 5, 20 elements of order 3, and 45 elements of order 2 or 4. Clearly this is impossible, hence two sylow groups of order 4, call them H and J, intersect in a subgroup V of order 2. Let K be the normalizer of V. K is not G, else V is normal. Since H and J are abelian, V is normal in H and J, and K contains both H and J. Therefore K is proper between H and G. It's index is at most 5, and that puts us back into A₅.

61. prime

62. third sylow 31

63. third sylow 7

64. prime power

65. third sylow 13

66. third sylow 11

67. prime

68. third sylow 17

69. third sylow 23

70. third sylow 7

71. prime

72. There are four instances of the sylow subgroup of order 9, and they are all conjugate. Thus the normalizer of such a subgroup has index 4 in G. Use this subgroup in the strong cayley theorem, and 72 does not divide into 4!.

73. prime

74. third sylow 37

75. third sylow 25

76. third sylow 19

77. third sylow 11

78. third sylow 13

79. prime

80. strong cayley 16

81. prime power

82. third sylow 41

83. prime

84. third sylow 7

85. third sylow 17

86. third sylow 43

87. third sylow 29

88. third sylow 11

89. prime

90. The sylow subgroup of order 9 has 10 conjugates in a simple group of order 90. If they are disjoint they consume 80 elements, leaving only 10 for the 6 subgroups of order 5. Thus at least two groups intersect. Let V be the intersection of H and J, as we did in order 60. Let K be the normalizer of V. K is not G, yet K contains both H and J. The index of K is at most 5, and that puts G into A₅, which is impossible.

91. third sylow 13

92. third sylow 23

93. third sylow 31

94. third sylow 47

95. third sylow 19

96. strong cayley 32

97. prime

98. third sylow 49

99. third sylow 11

100. third sylow 25

101. prime

102. third sylow 17

103. prime

104. third sylow 13

105. The group of order 105 has 15 groups of order 7, consuming 90 elements of order 7. At the same time we need 21 groups of order 5, and we can't accommodate that with the 15 elements remaining.

106. third sylow 53

107. prime

108. strong cayley 27

109. prime

110. third sylow 11

111. third sylow 37

112. strong cayley 16

113. prime

114. third sylow 19

115. third sylow 23

116. third sylow 29

117. third sylow 13

118. third sylow 59

119. third sylow 17

120. There are 6 sylow groups of order 5, hence G embeds in A₆. Since G does not fit into A₅, the action is transitive. K stabilizes 0, and K has order 20. Groups of order 20 have been characterized. A 4 cycle cannot live in 5 elements, since it would be an even permutation, nor a 10 cycle. Yet every group of order 20 has one or the other.

121. prime power

122. third sylow 61

123. third sylow 41

124. third sylow 31

125. prime power

126. third sylow 7

127. prime

128. prime power

129. third sylow 43

130. third sylow 13

131. prime

132. If G is simple, 12 sylow subgroups consume 120 elements of order 11, and 4 sylow subgroups consume 8 elements of order 3. This leaves just enough for the sylow subgroup of order 4, which is unique and normal.

133. third sylow 19

134. third sylow 67

135. third sylow 5

136. third sylow 17

137. prime

138. third sylow 23

139. prime

140. third sylow 7

141. third sylow 47

142. third sylow 71

143. third sylow 13

144. There are 16 subgroups of order 9. If these are disjoint they consume 16×8 = 128 elements, leaving 16 for the unique group of order 16. Therefore some of these groups intersect. Let H and J intersect in V, and argue as we did in order 60 and order 90, to create a normalizer whose index is at most 8. If the index is 8 then the order of the normalizer is 18. Such a group cannot have two sylow subgroups of order 9, hence the index is at most 4. That puts G into A₄, which is impossible.

145. third sylow 29

146. third sylow 73

147. third sylow 49

148. third sylow 37

149. prime

150. strong cayley 25

151. prime

152. third sylow 19

153. third sylow 17

154. third sylow 11

155. third sylow 31

156. third sylow 13

157. prime

158. third sylow 79

159. third sylow 53

160. strong cayley 32

161. third sylow 23

162. strong cayley 81

163. prime

164. third sylow 41

165. third sylow 11

166. third sylow 83

167. prime

168. There are 8 sylow groups of order 7, 28 groups of order 3, and 21 groups of order 8. Yes, 7 froups of order 8 are permitted, and that puts G into A₇. In fact there is a new simple group of 168, living in A₇, that will be described below. For now assume G does not fit into A₇, whence it has 21 groups of order 8. If these groups are disjoint they consume 147 elements, which does not leave room for 48 elements of order 5 and 56 elements of order 3. So H and J intersect in V. If V is normal in H then it has a proper normalizer between H and G, of index 7 or less. Remember that a group of order 4 is normal in H. If V is not normal then review the groups of order 8; V is a reflection in D₄. The rotations are unique to each group, consuming 3×21 = 63 elements. Add this to 48 and 56 and the identity to get 168. There isn't room for a single reflection in any of the 21 dihedral groups. Therefore G embeds in A₇.

     Let K stabilize 0, hence the order of K is 24. A group of order 8 lives in K, and consists of even permutations on 6 elements. An 8 cycle is impossible, and a 4 cycle shifts 4 elements and swaps the other 2. Suppose another involution commutes with this, to make an abelian group of order 8. A transposition cannot connect the 4 cycle and the 2 cycle. A transposition cannot align with 2 adjacent elements in the 4 cycle. Finally if a transposition swaps 1 and 3 in the first 4 elements then it becomes a circular shift of 2: 3412.

     Try to turn Z/4 into dihedral D₄, or quaternion Q₈. If the 4 cycle rotates the square, then the 180 degree rotation is in the center. If the 4 cycle is i, -1, -i, and 1, then again the circular shift of 2 is in the center. Look for a double transposition that commutes with 3412. A transposition cannot be both in and out of this cycle. A transposition wholly contained in the cycle does not commute with it either, unless it is one of the two transpositions that is already there. Relabel 1 through 4 if need be, and this last involution is 321465. This works, and produces D₄. The first 4 elements are the corners of a square, rotating (first generator) or flipping through a diagonal (second generator). We'll see later on that the second generator, that particular double transposition, combines with the 7 cycle to produce the simple group of order 168.

     The last group of order 8 is (Z/2)³. Again, to commute, transpositions coincide or are disjoint. The first generator is 3412, and the second is 563412. The last double transposition that aligns with these is already implied by them, 125634. That finishes off the groups of order 8. There is but one simple group of order 168, which we will explore below.

169. prime power

170. third sylow 17

171. third sylow 19

172. third sylow 43

173. prime

174. third sylow 29

175. third sylow 7

176. third sylow 11

177. third sylow 59

178. third sylow 89

179. prime

180. If G has order 180 it could have 6 copies of Z/5. G embeds in A₆, but then G has index 2 and is normal in A₆. Since A₆ is simple, this is impossible, hence there are 36 subgroups of order 5, consuming 144 elements.

     If there are four subgroups of order 9 then G embeds in A₄, which is impossible, hence there are 10 subgroups of order 9. If these are disjoint we need another 80 elements. At least two sylow groups intersect.

     Let H and J intersect in V, where V has size 3. Let K be the normalizer of V, so that K includes H and J. If K has order 18 then it cannot include two sylow subgroups of order 9. Thus the index of K is 5, 4, or 2, or K = G.

181. prime

182. third sylow 7

183. third sylow 61

184. third sylow 23

185. third sylow 37

186. third sylow 31

187. third sylow 17

188. third sylow 47

189. third sylow 7

190. third sylow 19

191. prime

192. strong cayley 64

193. prime

194. third sylow 97

195. third sylow 13

196. third sylow 49

197. prime

198. third sylow 11

199. prime

200. third sylow 25

This is a lemma, to help establish the simple group of order 168, though it can be applied to other groups as well. It is proved in 6 parts.

1. Let G be a permutation group within S_n. Assume the action is transitive, hence there is one orbit of size n. The stabilizer of any element is a subgroup of index n. All these stabilizers are conjugate, since they lie in the same orbit.

   Suppose a stabilizer contains K, a normal subgroup of G. Conjugate the stabilizer so that it fixes another element of S. Thus the conjugate of K, which equals K, fixes the second element. And the third, and the fourth, etc, hence K is trivial. If G is transitive, K does not live within a stabilizer.

2. Let G be transitive and let T be a block in S. Let H be the subgroup of G that maps the block T onto itself. Let J be the stabilizer of an element x in T. Since T is a block, J is wholly contained in H.

   Left cosets of J move x to other elements in T, and there is such a coset for everything in T, since G is transitive. Hence J has index |T| in H. This means |T| divides the index of J in G, which is n.

   When n is prime, all blocks have size 1 or n, and G is primitive. For n prime, G is primitive iff G is transitive.

3. Let G be a primitive permutation group on n letters. Suppose the stabilizer of x, call it J, is contained in a larger subgroup H. We will see that this cannot happen.

   Let T be the orbit of x with respect to H, so that H maps T onto itself. Since J is contained in H, each coset of J is entirely inside H or entirely outside H. The rest of G, outside of H, contains cosets of J that move x outside of T. That is the definition of T as an orbit of x under H. Now consider some other y in T. Something in H moves x to y, and the inverse moves y back to x. Suppose something outside of H moves z in T to y. Multiply by the element of H that moves y to x and find something that moves z to x. All in the same orbit, so the multiplier is in H, yet it is something in H times something outside of H. This is a contradiction, hence anything outside of H moves z out of T, and moves T out of T. Therefore T is a block. Since G is primitive, T is the single element x or all of S. There is no subgroup strictly between J and G.

4. Assume G is primitive on n letters, J is the stabilizer of x, and K is a normal subgroup of G. Recall that K is not contained in J (1), so K join J is properly larger than J, hence K join J = G (3).

   If R is K∩J, a dimensionality argument shows the index of R in K = the index of J in G = n. Therefore |K| is divisible by n.

5. Let G be a primitive group on p letters, where p is prime. The order of G divides p!, which has only one factor of p. Let K be a normal subgroup. By (4), p divides |K|. If x, outside of K, has order p, then the image of x in G/K has order p. Yet G/K is not divisible by p, so every element of order p lies in K. Every p cycle is contained in K. A normal subgroup contains all the p cycles.

6. Let G be transitive on p letters, and let K be a normal subgroup. Since G is transitive it is primitive (2). A normal subgroup K contains all the p cycles of G (5).

The simple group of order 168 can be described as a coliniation on 7 points. G carries triangles to triangles, rather than lines to lines. There are 7 triangles, and no two triangles share a side. Each triangle meets two others at each of its three vertices. If the points are a through g, think of the first triangle as efg, and connect these vertices to a b c and d as follows.

(efg) abe cde acf bdf adg bcg

Hold efg fixed, and you can swap ab and cd, or ad and bc, or ac and bd. Triangles map to triangles under these operations. This is a group Z/2 * Z/2 inside G.

Another symmetry exists around e f and g. Focus on e and select a different "primary" triangle, such as abe. The remaining four points are cdfg. Each of a b and e touches two other triangles, and still triangles never share a side.

Map efg onto abe, or any of the 7 triangles for that matter, then assign the vertices in 6 different ways. Or if you prefer, map e to any point, map f to any point connected to e, which is any of the remaining 6 points, then map g to the point that completes the triangle. No choice about the destination of g; there is only one triangle per side. You can map the primary triangle to another in 42 different ways.

The points abcd must map onto the points opposite the new triangle. There are 24 possibilities, but they don't all work. I'll illustrate by leaving efg in place. We already saw the group of order 4. Map a to any of a b c or d, and there are three triangles, with sides from a to e f and g, that must remain triangles once a has moved. This carries b c and d along. The destination of a determines everything else, and the group is Z/2 * Z/2. The order of g is 7×6×4 = 168.

This group is a permutation group on 7 letters. Its order is divisible by 7, so it has an element of order 7, which is a 7 cycle. Two examples are:

efdagcb

efbcgad

These permutations have order 7. In particular, the first permutation maps a to e g b f c d, and back to a. Both permutations carry a to e to g, and diverge from there, hence they are independent 7 cycles. Finally, verify that they map triangles to triangles; I'll leave that one to you. We can now prove the group is simple.

Since G is transitive and 7 is prime, apply the previous theorem part 6. A normal subgroup H of G contains all the elements of order 7. By the third Sylow theorem, the number of subgroups of order 7 is 1 mod 7, and divides 24. We produced two independent 7 cycles, so there must be 8 subgroups of order 7. These 8 groups define 48 elements of order 7, so |H| ≥ 49. Since |H| divides 168, it is either 84 or 56.

If |H| = 84, then H cannot contain 8 instances of Z/7 (third Sylow theorem). Thus |H| = 56.

Since H has 8 elements not of order 7, these form the Sylow subgroup of order 8 in H. This 8-group does not contain the 7 cycles of G, and is not normal in G, yet it is unique in H, which is normal in G. Conjugate H, and perform an automorphism on H, which maps the 8 sylow group onto itself, hence it is normal in G after all. This is a contradiction, thus the normal subgroup H cannot exist, and G is a simple group.

Sometimes G is presented as a permutation group on 7 letters, with a circular shift as a 7 cycle. Renumber our points so that a e g b f c d are designated 0 through 6. Recall that our first 7 cycle stepped through these points in this order, so this becomes a circular shift of 0 through 6. Then, swapping ab and cd becomes the double transposition on 03 and 56. Here are the generators of G168, without any reference to the underlying coliniation.

1 2 3 4 5 6 0

3 1 2 0 4 6 5

The rest of this chapter deals with n×n matrices over a finite field F, having characteristic p and order w, where w is some power of p. The matrices with nonzero determinant form a group, since each is invertible. This is the general linear group. This group maps onto F^* by the determinant, which is a homomorphism that respects multiplication. The kernel of this homomorphism is the group of matrices with determinant 1. This is the special linear group; I will simply call it G.

If the determinant of M must equal 1, then the center is c times the identity matrix, where cⁿ = 1. The 3 by 3 matrices mod 7 have a center of I, 2I, and 4I.

Let M be lower triangular, with ones on the main diagonal, and consider M^p. Let L be everything below the main diagonal, so that M = L+1. Since L and 1 commute, apply the binomial theorem. Everything inside drops out, leaving 1^p + L^p. Each successive power of L pushes the entries down and to the left. L² has nothing on the main diagonal and nothing on the subdiagonal. L³ is one stripe lower still. Finally Lⁿ = 0. If n ≤ p, then L^p = 0, and M^p = 1. Since p is prime, the order of M cannot be less than p, unless M is the identity matrix with order 1.

If n is greater than p, then raise M to the p². The reasoning is the same, just apply the frobenius homomorphism twice. In fact the order of m is the first power of p that is at least n.

The standard p sylow subgroup S has ones down the main diagonal and arbitrary entries below. Verify that S is a subgroup, closed under multiplication, and the determinant is always 1. The order of every matrix in S is a power of p, as described above. The size of S, w raised to the (n²-n)/2, agrees with the powers of w in the order of G, (Remember that W is the order of F.) So S is a sylow subgroup, and all other sylow subgroups are conjugate to S. That includes the reflection, the upper triangular matrices with ones on the diagonal.

If you want every matrix in S to have order p, then n need only be bounded by p, not w, which could be quite a bit larger than p.

Don'T assume the sylow subgroup is abelian. Let n = 3, and let A have ones on and below the main diagonal. Let B = A, except for 2 in the second row first column. Verify that AB ≠ BA.

What is the normalizer of S? Assume ZS/Z = S. Suppose Z has a nonzero entry l above the main diagonal, in row i column j. Select the least i for which this is the case. Thus Z is 0 above row i and to the right of the main diagonal. Expand Z times Z inverse = 1, and verify that Z inverse is also 0 above row i and to the right of the main diagonal.

Let Z conjugate the matrix with ones on the main diagonal and 1 in row j column i. Consider the identity and the extra 1 separately. The conjugate of the identity is the identity, so concentrate on the extra 1. Z times this matrix moves column j (in Z) over to column i, and leaves the rest of Z on the cuttingroom floor. The entry in i,i is now l. Multiply this by Z inverse and get various scale multiples of row i in Z inverse. In particular, the i^th row is multiplied by l. This has to come out 0 to the right of the main diagonal. Therefore Z inverse has to be 0 on the i^th row to the right of the main diagonal. Remember that Z inverse is already 0 above row i and to the right of the main diagonal. Put this all together and Z inverse is 0 on and above row i, and to the right of the main diagonal. The same must be true of Z. Yet Z_i,j is nonzero. This is a contradiction. Therefore every matrix that moves S onto itself is lower triangular.

Conversely, let Z be lower triangular. Separate Z, and Z inverse, and a matrix from S, into a diagonal piece and a lower piece. Expand the product to get 8 terms. Most wind up in the lower triangle. Only the product of the three diagonals lives in the main diagonal. Since diagonal entries in Z and Z inverse are inverses of each other, this product comes out all ones. The result is in S. The normalizer is lower triangular, with diagonal elements adjusted in any way you like, provided the determinant is 1. The size of the normalizer is the size of the sylow subgroup times (w-1)^n-1.

If D is the abelian group of diagonal matrices, then S and D belong to the normalizer, and their sizes are relatively prime, with the product equal to the size of the normalizer; hence they generate the normalizer. In fact D represents the normalizer mod S, giving a semidirect product.

The number of sylow subgroups is the index of the normalizer in G. The order of G was computed earlier. Divide out by the normalizer, and the number of sylow subbgroups is (w+1) × (w²+w+1) × (w³+w²+w+1) × … × (w^n-1+…+w²+w+1). This is 1 mod p, as it should be.

Let M be a matrix in G, build the characteristic polynomial c(x), and extend F up to K, where c(x) splits. Apply Schur's Theorem, and M is similar to a lower triangular matrix T with the eigen values on the diagonal. If these eigen values are all 1, T^p is the identity matrix. (This assumes n ≤ p; you might need a higher power of p if n is large.) Thus M^p = 1. Conversely, let one of the eigen values be e, where e ≠ 1. Note that p is relatively prime to the order of the multiplicative group of K. Raising to the p power permutes the elements of K. Everything has a unique p^th root, and only 1^p = 1. Thus e^p is something else. In other words, M^p is not 1. Therefore M is a p^th root of 1 iff its eigen values are all 1.

In general, the order of M is l, the least common multiple of the orders of its eigen values, or p times as much. Raise M to the l to get eigen values of 1; and no lesser exponent will do. This is then the identity matrix, or else it becomes so when raised to the p.

If M was diagonal then l is all we need. M^l is diagonal, with ones down the diagonal, hence 1. The same holds if M is similar to a diagonal matrix. However, if M is not diagonalizable then its jordan form has at least one nontrivial jordan block. Raise this to the l and the subdiagonal entry becomes l. Thus M^l is not 1. M has order l if it is diagonalizable, and order lp if it is not.

A normal subgroup of G contains all the nontrivial matrices with eigen values of 1, or none at all.

Start with the identity matrix and place an x somewhere below. This is called an atomic matrix. Multiply this matrix by itself again and again and get 2x, 3x, 4x, and so on. This is the integer multiples of x, but in fact every value of F is accessible. Scale this row by y/x, and the row below by x/y to keep the determinant 1. (Remember the conjugator must lie in G.) The corresponding columns are also scaled, but that doesn't mess with x; thus x turns into y. If x is on the floor, scale x and any row other than the column of x. If that is not possible, then n = 2. In this case, conjugate by a diagonal matrix with s in the upper left and t (s inverse) in the lower right, and x is multiplied by t². All the square multiples of x are accessible. In characteristic 2 we are done; so let p be odd. Multiply two of these modified matrices together and get x times (a²+b²). By the norm distribution theorem, two squares can sum to anything in F. Thus Every value of F is accessible.

If F is the reals then every positive real is a square, and you also get -x because the normal closure includes the inverses.

C is no trouble, because everything has a square root. The same holds for the algebraic closure of Q (bringing in -1 again), or the algebraic closure of a finite field.

Given an atomic matrix, conjugate by a transposition matrix that swaps rows i and j, and columns i and j. This puts the identity matrix back the way it was, but it also moves x to a new row or column. Try a few examples and see. Thus x can be moved anywhere below the main diagonal, and all these matrices are at our disposal.

Remember that the transposition matrix has to have a -1 somewhere, to keep the determinant 1. If x is negated that really doesn't matter, because x implies everything in F.

Now write any matrix in S as the product of these atomic matrices. Read the entries from top to bottom, left to right, and multiply the corresponding atomic matrices together, in that order, to realize the given matrix in S. Therefore, any atomic matrix implies all of S.

Given a matrix M in S, how do we isolate a single value x? If n = 2 then x is already alone, so assume n > 2.

If M is in S, convert M to jordan canonical form. The eigen values are all 1, hence you do not need to extend the base field F to make this conversion. In other words, the conjugator lives in the general linear group. If the conjugator has determinant 1/a, conjugate again by the diagonal [1,1,1…a]. Now the conjugator lives in G. Assume this has been done. The subdiagonal has scattered ones, and perhaps a on the floor.

If the subdiagonal is 0 then conjugate back and find that M was the identity after all. M is nontrivial, so the resulting jordan form has something nonzero on the subdiagonal.

The jordan blocks are readily apparent, you can read them off by traveling down the subdiagonal, but you can also subtract the identity and raise the resulting matrix to various powers, and watch jordan blocks go away. Even the bottom block, which may not be a proper block, having a on the floor; you can still subtract the diagonal and raise to the l, where l is the size of the block, and that block drops to 0. Raise to the l-1 to find a on the floor in column n-l, evidence of the block of size l that was once there.

Let the largest block have size l and consider the first (highest) of these blocks. Assume l is at least 3. For notational convenience, assume this happens at the top, with ones in positions 2,1 and 3,2. Negate rows 2 and 3, and columns 2 and 3. (You can skip this step if p = 2.) That negates the first entry in the subdiagonal and leaves the second one alone. The third entry is also negated (if it was nonzero). Multiply by the original matrix, and the first entry goes away, while the second entry is doubled. The third entry goes away, if it was nonzero before. But now there are entries on the next stripe down, just below the subdiagonal - entries for this block and for every other block of size at least 3.

What has happened to the sizes of the various jordan blocks? To find out, subtract the identity and raise each block to various powers. Each lesser block, i.e. every block that we did not mess with, has its subdiagonal doubled. The block is the same size, or less if p = 2. The block that we are meddling with is a bit different however, because the first entry on the subdiagonal is gone, whether p = 2 or not. If it was size l before, it is size l-1 now, or less. Yet the block is not gone entirely, because the subsubdiagonal remains. Thus one of the largest blocks has been reduced in size, and the other blocks are the same or smaller.

Convert back to jordan form and do this again, and again, untill all the blocks have size 2 or less.

If the subdiagonal has a single nonzero entry, call it x, and we're done.

Perhaps there are still multiple nonzero entries on the subdiagonal. Thus n is at least 4; assume n is greater than 4. Assume p is odd, so that we can meaningfully negate rows and columns. Negate two rows, and two columns, so that the first nonzero entry on the subdiagonal is negated, while the second one is untouched. Multiply by the original, as we did before. This kills off the first entry in the subdiagonal, and doubles all the rest. Repeat this process, moving down the line, until the matrix becomes atomic.

Now let p = 2. Assume there is some wiggle room on the subdiagonal. There are enough zeros that you could slide a 1 up or down the subdiagonal, and it would still have zeros on either side. Permute three rows, and three colums, to swap the block of size 2 with the neighboring block of size 1, effectively sliding the 1 along the subdiagonal. Multiply this by the original. The result is a block of size 3, while all the other blocks go away. Square this, to create a block of size 2, and you're done.

The last case is n = 4, with subdiagonal [1,0,a]. Scale the last two rows by s and t, where st = 1, and the last two columns by t and s. This multiplies a by t². If p = 2, and w > 2, let t be anything other than 1. Multiply this matrix by the original and the first block goes away, leaving a*(1+t²) in position 4,3.

If p is odd, and w > 3, let t be anything other than 1 or -1. This modifies a, and leaves 1 alone. Multiply this by the inverse of the original. Once again the first block goes away, leaving an atomic matrix with something nonzero in position 4,3.

Any lower triangular matrix with eigen values of 1 has, in its normal closure, an atomic matrix, which then brings in all of S.

Restrict to a subfield of F, and the matrices restrict to a subgroup of G, but not a normal subgroup. As shown above, any atomic matrix brings in S with entries in all of F.

Instead of triangular matrices, let's talk about eigen values. If M has all eigen values of 1, convert M to a lower triangular matrix by Schur's theorem. Do this without leaving F, because all eigen values are 1. Adjust the conjugator to have determinant 1, which still keeps the result lower triangular. This brings in all of S, and it brings in every other matrix with eigen values of 1, since each such matrix is similar to something in S by Schur's theorem, and still similar to something in S when the conjugator is adjusted to have determinant 1. As the title of this section proclaims, a normal subgroup of G has no matrices with eigen values of 1, or it has them all.

If F has characteristic p and p is at least n, the same statement can be made about p cycles, because M (not the identity) is a p cycle iff its eigen values are 1. A normal subgroup has one p cycle iff it has them all. Rephrase this as a power of p if n is larger than p.

For the rest of this chapter, an eigen value is called "real" if it lives in F. An imaginary eigen value lives outside of F, and requires a field extension. This is analogous to real and imaginary numbers, though the analogy is not perfect.

If a normal subgroup of G includes S, then it also includes the matrix with [2,s] on top, and [t,0] below, where st = -1. For n > 2, the rest of the matrix can be the identity. When p = 2, the upper left is 0. Verify the eigen values are all 1.

Multiply this by the atomic matrix with x in position 2,1. By adjusting x, the upper left, 2+sx, can be anything you like. Start with s and t, then select x so that the upper left comes out 1.

Do the same for u and v, where uv = -1, then multiply the two matrices together to get the following matrix on the left. Add s times the second row to the first, then add a multiple of the first row to the second.

The third matrix is diagonal, and the entries are eigen values. Since s and v are unrelated, any eigen value is fair game. The "other" eigen value takes up the slack, giving a determinant of 1.

Apply the above reasoning to the second and third entries, and build a diagonal matrix whose entries are all ones, except for the second entry, which is arbitrary, and the third entry, which is the inverse of the second. Do the same for the third and fourth entries, and the fourth and fifth, and so on. These diagonal matrices can be multiplied to produce any diagonal matrix in G.

The abelian diagonal group combines with the sylow subgroup S to generate the normalizer N, i.e. all the lower triangular matrices. In summary, one p cycle is enough to bring in all of N, and all the conjugates of N follow from there. This includes every matrix with eigen values in F, as these are similar to a lower triangular by Schur's theorem. All this comes from one matrix with eigen values of 1.

Conversely, start with any matrix whose eigen values lie in F, and convert it to jordan canonical form. The conversion does not require an extension of F, hence conjugation occurs within our group. The result is diagonal with scattered ones on the subdiagonal. Again, the entry on the floor might be a instead of 1, so that the conjugator has determinant 1. Let the diagonal have order l. This is the least common multiple of the orders of the eigen values. (For the first time in this proof, F really needs to be finite.) If l is 1 then all the eigen values are 1. We either have 1 or a p cycle. Only the identity matrix is similar to the identity matrix, so if the initial matrix was nontrivial, then we have a p cycle, and everything follows from there.

Next assume l is not 1. There is at least one eigen value that is not one,yet they all lie in F. If there is a jordan block with a subdiagonal, let d be the diagonal and let y be the subdiagonal, and raise d+y to the l. Look at the top two diagonal stripes of this block. Since d and y commute, use the binomial theorem. The diagonal turns to 1, and the subdiagonal is ld^l-1y. This is 0 only if l is 0. Yet l is w-1, or a factor thereof. The jordan matrix, raised to the l, has something just below the main diagonal, and is a p cycle.

Finally assume the matrix is diagonal. If it is constant then it is in the center of G. That is a normal subgroup all by itself, and I'll assume this has been pulled out. We're really looking at G mod its center.

Assume the diagonal is not constant. Permute the entries so that the same eigen values are grouped together. Thus the eigen value at the upper left is not equal to the eigen value at the lower right. Call these s and t respectively. Premultiply by an elementary matrix that adds the first row to the last. Then post multiply by the inverse, which subtracts the last column from the first. The lower left corner is now s-t, which is nonzero. Call it x.

Raise this matrix to the w-1. The diagonal becomes 1, giving 1 or a p cycle. It all depends on the lower left corner. Things don't commute, so you can't use the binomial theorem, but you can expand (d+x)^w-1 by the distributive property, and throw away any term with two instances of x, as these are all 0. This leaves the sum of dⁱxd^w-1-i, as i runs from 0 to w-1. This in turn is the sum of tⁱxs^w-1-i. Pull x out, and the remaining factor is (s^w-t^w)/(s-t). (Since s ≠ t, this is well defined.) Since s^w = s, and t^w = t, the expression is 1. Therefore x survives, and the matrix is a p cycle.

In summary, a normal subgroup of G mod its center contains every matrix with eigen values in F, or it contains none of them.

If a normal subgroup H of G contains a p cycle, then all the p sylow subgroups are in, along with their normalizers, as described above. Let z be the size of the normalizer of a sylow subgroup. Within H, each sylow subgroup remains a sylow subgroup, and H contains all of them. The standard sylow subgroup S still has its normalizer of size z, and the number of sylow subgroups, call it k, has not changed. Therefore |H| = z*k (third sylow theorem). This is all of G. In summary, a normal subgroup that contains any matrix outside the center with real eigen values, becomes all of G.

Apply this to G mod its center. Start with a normal subgroup in G/C that contains a matrix with real eigen values. Real eigen values remain real when you multiply by a constant diagonal matrix in the center, so this is well defined. Lift (by correspondence) to a normal subgroup of G. This contains the aforementioned matrix, and becomes all of G. Hence the normal subgroup downstairs is G/C. If we can get from an arbitrary matrix to a matrix with real eigen values in F, then G/C is simple.

Recall that the size of the group of 2 by 2 matrices with determinant 1 is w*(w²-1). If w is odd then the center is ±1, and we must divide this count by 2 to get the size of G mod its center. For w = 2, |G| = 6, and for w = 3, |G/C| = 12. Clearly these are not simple groups. However, when w = 4, |G| = 60, and when w = 5, |G/C| is also 60. These are isomorphic to A₅, the only simple group of order 60. Similarly, w = 7 gives the simple group of order 168. Set w = 8 for 504, and w = 9 for 360. Like the prime cycles and the alternating groups, this is another class of finite simple groups.

The goal of this section, and the next, is to multiply various conjugates of M together to produce a triangular matrix, which always has eigen values in F. To be a bit more general, let M be nonsingular, with determinant y; though the conjugators, that lead to a triangular matrix, will always have determinant 1. Set y = 1 to get back into G.

In characteristic 2, the lower left is always a square, for any s. For odd p, suppose the lower left is never a square. There are (w-1)/2 nonsquares, and each admits at most 2 values of s by the quadratic formula, so some s leads to a square in the lower left. If this square is 0, then the matrix is upper triangular, and not central, since b is nonzero. (The diagonal entries are nonzero because the determinant of this matrix is still y.) So there is a nonzero square in the lower left.

In SL₂(F), y = 1, but even if we did not know y was 1, it is definitely nonzero in F. This matrix has real eigen values, namely -y. If the lower left is nonzero, then we are not in the center, and all the matrices with real eigen values come rolling in, along with everything else. We can stop here, unless p = 2, or a+d = 0.

Multiply two such matrices together, based on t and u, where t ≠ u. The upper right is nonzero. How about the lower left? The expression simplifies to (u-t)*(y-ut). Select any u and t so that their product is y, and the result is upper triangular. That completes the case a+d = 0, unless this can only be done when u = t. Write t² = y, and there are but 2 solutions. As long as there are more than 2 nonzero values in F, there is no trouble. We only need look at p = 2 and p = 3. Let's look at these now.

When p = 2, G has order 6, and is nonabelian. Such a group is a semidirect product of Z/3 by Z/2, and there is only one such group, namely S₃.

When p = 3, G mod ±1 has order 12, and cannot be simple. Swap rows and columns, and S, the standard lower triangular group of order 3, becomes upper triangular, hence S is not normal in G. With 4 sylow subgroups of order 3, there are 8 elements of order 3. The remaining 4 elements form the single, normal sylow subgroup of order 4. Multiply these by each other, negating the result any time you wish, and this subgroup is Z/2 * Z/2.

To confirm our earlier results, note that the antidiagonal matrix above has no eigen vectors, and no real eigen values.

To find the normalizer of S, change the diagonal; but the determinant must be 1, so that leaves [1,1] or [2,2]. These are the same mod the center, hence the normalizer of S is S.

Let G/C act on sylow subgroups by conjugation, and since the normalizer is no larger than S, the group embeds in A₄. G/C already has size 12, so it is isomorphic to A₄.

Don't confuse G with S₄. Both have order 24, but G has a center of Z/2 and a quotient of A₄, while S₄ has a normal subgroup of A₄ and a quotient of Z/2.

If F is a finite field with order > 3, and G = SL₂(F), then G mod ±1 is a simple group.

Recall that G/C is isomorphic to the quaternions with norm 1, mod ±1, hence this quaternion group is also simple. At the same time, these two groups are isomorphic to the 3×3 orthonormal matrices that are squares, hence that group is also simple. Either of these groups, quaternions or orthonormals, would be difficult to analyze from first principles.

When n = 2, the special linear group, mod its center, is simple (except for 2 and 3). This was proved in the previous section. Proceed by induction on n.

Actually I'm going to prove something a bit more general. Given a noncentral matrix M with a nonzero determinant, there is a product of conjugates of M that is lower triangular and noncentral. The conjugators will always have determinant 1. As a corollary, apply this result when det(M) = 1, and find a lower triangular matrix, with determinant 1, that is noncentral in G, and nontrivial in G/C. This has real eigen values, and brings in the entire group, as shown above. Therefore G/C is simple.

In the 2 by 2 case, I was careful to show that any matrix leads to a lower triangular matrix, as long as its determinant is nonzero. So we can take the inductive step, as long as F is not 2 or 3. I'll address that as needed.

Start with any matrix M that is noncentral. We want a 0 in the upper right. If it is not already zero, look at the entry below. If that one is 0 then swap the first and second rows, and first and second columns, to put 0 in the upper right. (This swap is conjugation by an elementary matrix. It also negates one of the rows / columns, but that doesn't matter.) If the top two entries on the right are both nonzero, use an elementary matrix to subtract a multiple of the second row from the first, while adding a multiple of the first column to the second. This makes the upper right 0.

Next, make the entry in row 2 column n 0. Do this in exactly the same way, using rows 2 and 3. Continue this all the way down to row n-2.

Zero out column n-1, from the top down to row n-3. Zero out column n-2, from the top down to row n-4. Continue all the way over to the entry in row 1 column 3. Now M is almost lower triangular, except it has entries just above the main diagonal. I will call this the superdiagonal, as opposed to the subdiagonal which lives below the main diagonal.

So far the operations have all been conjugation by elementary matrices. The new matrix cannot be central, because the original was not central.

Assume the upper left block A is not lower triangular. By induction, products of conjugates of A lead to a lower triangular matrix that is not in the center. Whenever you conjugate A by X, conjugate M by a matrix that is X in the upper left and diagonal 1 in the lower right. The determinant of this larger X containing matrix is still 1. Then multiply these conjugates together as needed. The result is a lower triangular version of A in the upper left, and some rows below that we don't know much about. The determinant is still nonzero, hence the determinant of D is nonzero.

Since the converted form of A is noncentral, M is noncentral. If D happens to come out lower triangular then we're done. Otherwise, by induction, a product of conjugates of D will create a lower triangular nonsingular block. Whenever you conjugate D by X, conjugate M by a matrix that is X in the lower right and diagonal 1 in the upper left. Then multiply these conjugates together as needed. The result is the lower triangular version of D in the lower right, a lower triangular matrix in the upper left, and some stuff in the lower left that we don't care about. A might come out central, by accident, while converting D, but D is not central, so M is not central. That completes the conversion of M, except for the troublesome case of F = 2 or 3, with one or both blocks having size 2. I'll address that now.

For F = 3 and n = 2, matrices of order 3 are conjugate to something lower triangular, so assume A is any of the matrices in the normal subgroup of order 4. It is not 1, as that is already lower triangular. If it is antidiagonal then leave it alone. If A is either of the other two, conjugate to get the antidiagonal matrix [0,1] [-1,0]. Square this to get -1. This is central in A; we have to make sure it doesn't become central in M.

Assume n = 3. Thus D, of size 1, is already central. M becomes central only if D² = -1, and that can't happen mod 3.

Assume n = 4, and blocks A and D have size 2. The blocks become triangular or antidiagonal by pure conjugation. Since the original matrix was not central, the conjugated version is not central either. If both blocks are triangular we are done. So assume A is antidiagonal. If D is lower triangular then square the matrix, and A = -1, and the diagonal of D = 1. This is noncentral. Therefore both A and D come out antidiagonal. Square this and the diagonal is -1. We need M² to have something nonzero in C. Suppose it comes out 0. Put four unknowns in the C block and solve four equations in four unknowns. Depending on the distribution of 1 and -1, the matrix looks like this. Square it and see.

Add row 2 to row 3, and subtract column 3 from column 2. This tweaks x by 1. Multiply this matrix by the original, and get x from the first matrix, and x+1 from the second, which leaves something nonzero in C. The product has -1 down the diagonal, and stuff below, and is noncentral. That takes care of n = 4.

Finally let n be 5 or more, where A has size 2. Let D be triangular, or having been made so. If A is triangular then we are done. If A can be conjugated triangular then we are done. So assume A has become antidiagonal. Square M, and as before, A becomes -1, while the diagonal elements of D are all 1. Thus M becomes noncentral and triangular, and that completes the special case of F = 3.

The case of F = 2 will not be dealt with here. Perhaps a later edition of this book.

Next, assume the superdiagonal is entirely nonzero. M cannot be cut up into four blocks. We want a 0 in the upper left. Since the entry nextdoor is nonzero, use an elementary matrix to add a multiple of the first row to the second, and subtract a multiple of the second column from the first, thus zeroing out the upper left. In fact you can zero out the entire main diagonal, except the last entry in the lower right.

With 0 in the upper left, we want a 0 below. Use an elementary matrix to add a multiple of the first row to the third, and subtract a multiple of the third column from the first, thus zeroing out that entry. Continue down the line, and zero out the entire subdiagonal, except the last entry on the floor. Repeat this process until M has only a bottom row and a superdiagonal.

Conjugate M by an antidiagonal of ones, with -1 in the lower left if necessary to make the determinant 1. This rotates M 180 degrees, and then (perhaps) negates the bottom row and rightmost column.

Multiply the two matrices together and get a diagonal plus some stuff on the bottom row. The first n-1 entries on the diagonal are nonzero because the superdiagonal was nonzero. The last entry has to be nonzero because the determinant is nonzero. If this lower triangular matrix is not in the center, then we are done.

If the first matrix is M, then the lower right entry in the third matrix, i.e. the lower triangular matrix, is M_n,1². Thus M_n,1 is nonzero.

Let p be odd, and look at the diagonal of the product. If it is nonconstant, then the product matrix is not central. If it is constant, let's tweak it. The first entry is M_1,2M_n-1,n. The i^th entry, for i < n, is M_i,i+1M_n-i,n-i+1. Negate rows 1 and 2, and columns 1 and 2, of M, then rotate and multiply as before. This negates something on the main diagonal of the product (unless n = 4), but not the entry in the lower right, which is a square. If n = 4, negate the first two rows and columns of M as before, but multiply by a rotated version of the original. This negates the last entry in the main diagonal but not the first.

If p = 2 and F > 2, select s and t such that st = 1 and s ≠ t. Scale rows 1 and 2 of M by s and t respectively, and scale columns 1 and 2 by t and s. The last entry in the main diagonal is multiplied by t². For n > 3, the first entry in the main diagonal is multiplied by s^2. In characteristic 2, s ≠ t implies s² ≠ t², so the diagonal becomes nonconstant. For n = 3, scale the first and third rows, and first and third columns. This multiplies the first and second entries in the main diagonal by s², and leaves the third one alone. In each case the product matrix is lower triangular, and with some modifications, it is noncentral.

As mentioned earlier, the case of F = 2 is not addressed here. If F is a finite field other than Z/2, the special linear group over F, mod the center, is a simple group, except for n = 2 and F = 3. This is a class of simple groups, like Z/p, and A_n.

The algebraic closure of F also works, because any noncentral matrix M lives within a finite subfield, and brings in all of that subfield, which brings in all of F. This is an infinite simple group.