• Subnormal Series, Solvable Group
• Refinement
• Composition Series
• Subgroup of a Solvable Group
• Commutator subgroup, Abelianization
• Commutator subgroups and Solvability
• Central Series, Nilpotent
• The Center and a Normal Subgroup Intersect
• Finite Nilpotent Group
• The Intersection of Maximal Subgroups is Nilpotent
• Zassenhaus
• Jordan Holder
• Uniserial

Let's look at some nonabelian solvable groups. The dihedral group D_n represents the reflections and rotations of the regular n-gon. The rotations alone form the cyclic subgroup Z/n. This is a normal subgroup of index 2, giving the series {D_n, Z/n, 1}, having factor groups Z/2 and Z/n. Every dihedral group is solvable.

The symmetric groups S₂, S₃, and S₄ are solvable. The latter has the normal series S₄, A₄, [1234, 2143, 3412, 4321], and 1, with factors Z/2, Z/3, and Z/2*Z/2.

A finite simple group other than Z/p is nonabelian, and not solvable. For instance, A_n for n ≥ 5 is alternating simple and not solvable.

For n ≥ 5, there is but one normal subgroup of S_n, which is A_n, hence S_n is not solvable either.

Verify the following for the direct product of groups.

The direct product of arbitrarily many parallel subnormal series is subnormal. If one series is longer than the other, just fill in with 1. If we are taking an infinite direct product, there must be a bound on the lengths of the individual subnormal series, else the product series will not reach 1 in a finite number of steps.

The direct product of arbitrarily many parallel normal series is normal. At any level, the direct product of subgroups, normal within their component groups, remains normal in the product.

At any level, The factor group associated with the direct product of our normal series is equal to the direct product of the individual factor groups.

The direct product of abelian groups is abelian.

The direct product of solvable series is solvable.

The direct product of finitely many solvable groups is solvable. An infinite number of groups also works, unless the solvable series associated with the individual groups increase in length without bound.

You might be wondering why these groups are called solvable. The answer is based on galois theory. In short, a polynomial extends the base field, and that extension has a galois group, and if that galois group is solvable then the polynomial is solvable via an algebraic formula similar to the quadratic formula. This will become clearer in the next chapter.

A refinement is always possible whenever the factor group N_i/N_i+1 is not simple. A normal subgroup in the factor group pulls back to a normal subgroup of N_i. This is a consequence of the correspondence theorem.

If the factor group is abelian, any subgroup is normal and allows for a refinement. The two new factor groups on either side of the refinement are abelian, hence a refinement of a solvable series remains solvable.

If a factor group is finite abelian, separate it into a product of prime power cycles. Any subgroup allows for a refinement, so refine the series until all the factor groups become prime power cycles, and then prime cycles. If G is finite abelian then it admits prime cycles all the way down.

A subnormal series with simple factor groups is a composition series. It cannot be refined any further. Ironically, it is also called a decomposition series - the way information can be valuable and invaluable at the same time.

Every finite group has a composition series. Let N_i+1 be a maximal proper normal subgroup in N_i. If the factor group had a normal subgroup, it would pull back, by correspondence, to a larger normal subgroup in N_i.

A finite solvable group always has a solvable series with prime cyclic factors.

Like an integer factoring into primes, a composition series is unique up to order. In a modest example, Z/15 has a normal subgroup Z/5 with a factor group of Z/3, or it has a normal subgroup Z/3 with a factor group of Z/5. There is no other choice.

Sometimes the order is prescribed. S₇ has a normal subgroup of A₇ and a factor group of Z/2, and nothing else. However, if S₇ had any other normal subgroup, (which it doesn't), it would have to be Z/2, with a factor group of A₇. I will prove unique factorization into simple groups at the end of this chapter. You could prove it right now for abelian groups; it is a consequence of the unique factorization of the order of G as an integer. Each factor group, all the way down, is a prime dividing the order of G.

The subgroup of a solvable group is solvable.

Let G be a solvable group with series N, and let H be a subgroup of G. Intersect N_i with H to create a new descending chain of subgroups. Conjugate H∩N_i+1 by anything in H∩N_i, and find a subgroup that is contained in N_i+1 and also in H. The restriction to H always produces normal subgroups, hence we have a subnormal series for H.

This series is just as long as the original; shorter if two groups in the chain happen to coincide. Intersect with 1, and every group in the series is 1.

A homomorphism F takes N_i onto a quotient group Q, with kernel N_i+1. By the correspondence theorem, a subgroup in the domain produces a subgroup in the range, so restrict to H, and F produces a subgroup of Q. If Q is abelian then every subgroup of Q is also abelian. The factor groups of our new, restricted series are all abelian. If G is solvable then H is solvable.

Next consider the image of a solvable group G, under a homomorphism F. Let the image be Q, and let Q_i be the image of N_i under F. The image of a normal subgroup is normal, so each Q_i+1 is normal in Q_i. We have a series for Q that is, once again, no longer than the original.

Remember that the quotient N_i/N_i+1 is abelian. If A and B are cosets of N_i+1 in N_i, AB = BA. Applying F shows that the cosets of F(N_i+1) commute as well, and the factor group Q_i/Q_i+1 is abelian.

In summary, the subgroup or homomorphic image of a solvable group is solvable. As it turns out, the converse is also true.

Let K be normal in G, with factor group H, and assume H and K are solvable. Find a solvable series for H and take the preimage under F. This gives a descending chain of subgroups in G, each normal in the previous. The last subgroup is K.

Next, append the solvable series for K. This takes us all the way down to 1. We have a well defined series, but are all the factor groups abelian?

Certainly the series for K meets our criteria, so consider the preimages of H. Let J_i and J_i+1 be the preimages of H_i and H_i+1 respectively. By correspondence under F, J_i/J_i+1 is isomorphic to H_i/H_i+1. Since the latter is abelian, so is the former. Therefore G is solvable.

Every finite p group is solvable. Proceed by induction on the size of the group.

Let G be a p group with nontrivial center C. The center is solvable, and the quotient group is a smaller p group, hence solvable. .Therefore G is solvable.

Let G be a group, and let G′ denote the commutator subgroup of G, which is defined as follows.

For any x and y in G, xy/x/y is a commutator, and the group generated by these elements is the commutator subgroup. If x and y commute, xy/x/y = 1. If G is abelian, G′ is trivial.

Let q be any element of G, and conjugate the commutator xy/x/y by q. The result is the commutator q(xy/x/y)/q/(xy/x/y) * (xy/x/y), which is in G′. Thus conjugation by q, for any q in G, maps G′ onto itself, and G′ is normal in G.

Seen another way, any automorphism moves commutators to other commutators, hence G′ is preserved by every automorphism. We say G′ is "fully invariant" in G. Restrict automorphisms to inner automorphisms, and G′ is normal in G.

Let q and r represent cosets of G′ in G. We know that qr/q/r is a commutator, a member of G′. Write qr/q/r = w, for w in G′. Hence qr = wrq. Thus qr and rq represent the same coset of G′. The quotient group G/G′ is abelian. This is called the abelianization of G.

Conversely, assume G/H is abelian, and reverse the above argument. Write Hqr = Hrq, or Hqr/q/r = H, or qr/q/r ∈ H. Every commutator belongs to H, and H contains G′. The quotient of G is abelian iff the kernel contains G′.

Let x and y belong to the same conjugacy class of G. Thus y = wx/w for some w. With wx/w/x in G′, y/x is in G′. This means y and x are in the same coset of G′, and represent the same element in the quotient group G/G′. Each conjugacy class maps, in its entirety, to a unique element in the abelianization of G. However, this map is not injective. Separate conjugacy classes can map to the same thing in G/G′. In other words, the conjugacy classes do not define the quotient group. Consider S₃, the symmetric group on 3 letters. Each commutator is an even permutation, and G is nonabelian, hence G′ = A₃. The quotient has order 2, but there are three conjugacy classes: 1, the two circular shifts, and the three transpositions. Three conjugacy classes cannot correspond to a quotient group of order 2. In fact the first two conjugacy classes map to A₃, since they already live there, while the third class maps to the other coset of A₃ in S₃.

This generalizes to any nonabelian group G. The commutator subgroup G′ is nontrivial, and 1 is its own conjugacy class, so some other conjugacy class belongs to the kernel G′. Conjugacy classes correspond to G/G′ iff G is abelian.

Here is a brief excursion into category theory, which you may skip if you wish. The abelianization of G is universal. That is, any homomorphism f from G into an abelian group H is the composition of f₁ from G onto G/G′, composed with f₂ from G′ into H, and f₂ is unique.

Just as f′′ is the derivative of f′, so G′′ is the commutator subgroup of G′. You can continue taking commutator subgroups, just as you might take third and fourth derivatives etc.

Of course G′′ is normal in G′, with an abelian quotient group. As it turns out, G′′ is also normal in G. As described in the previous section, any automorphism of G permutes the commutators, which are the generators of G′. This in turn induces an automorphism on G′. The automorphism of G′ then maps commutators to commutators, which are the generators of G′′. Thus every automorphism of G induces an automorphism on G′′, and G′′ is normal in G. By induction this continues down the chain.

If the k^th commutator subgroup equals 1, G has a normal series, and G is solvable.

Conversely, assume G is solvable. We want to show the k^th order commutator subgroup is 1. This is true if the series consists of G and 1, whence G is already abelian. Proceed by induction on the length of the solvable series.

Let N₀ = G, and let N₁ be the first proper normal subgroup in the series. The series ends at N_k = 1. Since N₀/N₁ is abelian, N₁ contains G′. Restrict N₁, and the rest of the series, to G′, to find a smaller solvable group. By induction on the length of the series, the commutator subgroup of G′ drops to 1 after k-1 iterations. Step back to G, and its commutator subgroup drops to 1 after k iterations.

A group is solvable iff it has a finite chain of commutator subgroups. Since each subgroup is normal in G, this is a normal series, not just a subnormal series.

The commutator chain could be shorter than the subnormal series, e.g. when G is abelian and the series is a composition series. In practice, the composition series is more interesting. It describes the group, the way prime factorization might describe an integer. Of course the normal series defined by the commutator subgroups can be refined into a composition series. We'll see below that the composition series is the same no matter how you get there.

Apply this to G = Q₂₄, the quaternion group of order 24. Since Q₈ is normal, with ann abelian quotient, Q₈ contains G′. The commutator hi/h/i is j/i, or k. By symmetry, i and j are also generators, hence G′ = Q₈. Any other maximal normal subgroup of this solvable group has to contain G′, so Q₈ is it. That is the only way down the composition series.

The next abelianization has generators like ij/i/j = k² = -1, hence G′′ = ±1. The next subgroup down the composition series contains ±1, and has order 4. Let i generate Z/4, though j or k would do just as well. Put ±1 below, and then 1, and that completes the series, a refinement of the commutator series, and the only (up to isomorphism) complete subnormal series for G. However, Z/4 is not normal in G, since conjugation by h cycles i j and k around. There is no complete, normal series for G, even though G is solvable.

A subnormal series is a chain of descending subgroups; a central series is a chain of descending quotient groups, each time modding out by the center of the previous.

Let C₀ = G, and let C₁ = G mod the center of G. If G is abelian we are done, but if not, continue in this manner. Let C₂ = C₁ mod the center of C₁, and so on. Thus the central series is actually a series of factor groups.

If the central series is finite, G is nilpotent. An abelian group is nilpotent. Since a p group has a nontrivial center, every finite p group is nilpotent.

The finite direct product of nilpotent groups is nilpotent. The composite central series is the direct product of the individual central series. An infinite direct product may be nilpotent as well, if the lengths of the individual central series are bounded.

The quotient of a nilpotent group is nilpotent. Apply a homomorphism f(G) = H. By induction on the length of the central series, the theorem holds for G/C, where C is the center. Let K be the kernel of f. The cosets of K that have representatives in C become elements in the center of H. In other words, C maps into D, where D is the center of H. Let J be the join of C and K, which is the same as C cross K mod C∩K. Now G/J maps onto H/D. Since G/C maps onto G/J , (J containing C), G/C maps onto H/D. Therefore, H mod its center is the homomorphic image of G mod its center, and is nilpotent by induction. This makes H nilpotent.

If H is a subgroup of G, H is nilpotent. By induction, assume every subgroup of G/C (C is the center) is nilpotent. Let H intersect C in D. Each coset of C in G is now cut down to a coset of D. If there were two such cosets in one coset of C, that would imply two cosets of D in C, which is impossible. Furthermore, the cosets of D obey the rules of G/C, just like the cosets of C. In other words, we have a representation of G/C, using cosets of D. However, some of these cosets may be missing, i.e. the cosets that are not in H. The quotient H/D is a subgroup of G/C, and is nilpotent by induction. Now, the center of H could be larger than D, creating a quotient of H/D, which is also nilpotent. Since H mod its center is nilpotent, H is nilpotent.

Given a central series for G, build a corresponding normal series, in reverse order, as follows.

A trivial homomorphism maps G onto G, with kernel 1. Thus C₀ = G, and N₀ = 1.

Moving to C₁, a homomorphism maps G onto C₁, and its kernel is the center of G. Let N₁ be this kernel, the center of G.

A second homomorphism maps C₁ onto C₂. Combine this with the first homomorphism to get a map from G onto C₂. Let N₂ be the kernel of this map. Continue all the way to C_k = 1, whence N_k = G.

Since each N_i is the kernel of a homomorphism, each N_i is normal in G. Restrict G to N_i+1, and N_i is normal in N_i+1, giving a normal series for G running in reverse order.

The factor group N_i+1/N_i is the center of C_i, which is abelian. All factor groups are abelian, the series is solvable, and G is solvable. Every nilpotent group is solvable.

However, there are plenty of solvable groups that are not nilpotent. S₃ has no center, and no central series, even though it has an abelian subgroup Z/3 with an abelian factor group Z/2.

The center of a nilpotent group G, and any normal subgroup of G, intersect in more than just 1.

If the central series contains only G and 1, then G is abelian. Every normal subgroup intersects the center of G, which is all of G.

Proceed by induction on the length of the central series.

Let C be the center of G and suppose K is a disjoint normal subgroup. Let D be the direct product of C and K. Since C commutes with everything, the subgroup D, generated by C and K, is indeed the direct product of C and K.

Take x in C and y in K, and conjugate xy by z, for any z in G. This yields x times the conjugate of y. Since K is normal the conjugate of y lies in K, and the conjugate of xy lies in D, hence D is normal in G.

Mod out by C, and D maps to a normal subgroup of G/C. In fact D maps to a copy of K. More than that, K maps 1 for 1 onto a copy of K, which remains normal in G/C.

Since G is nilpotent, G/C has a nontrivial center. By induction, some element y is in the center of G/C, and in the image of K. Returning to g, y ∈ K represents a coset of C.

Select any x in G, and xy/x appears somewhere in K. Map this forward, from G into G/C. Since y winds up in the center, xy/x becomes y in the image. Thus xy/x is in the same coset as y. Write xy/x = wy, where w is in the center of G. But remember, xy/x is in K, which is 1 cross K in the direct product. Both components must agree. Therefore w = 1, and xy = yx. This holds for all x, so y is in the center after all. This contradicts the assumption that K and C are disjoint. Therefore every normal subgroup intersects the center of a nilpotent group.

This is a rather surprising theorem. A finite group G is nilpotent iff all its sylow subgroups are normal, iff G is the direct product of its sylow subgroups.

Let G be nilpotent and let H be a proper subgroup of G. We will show that the normalizer of H is larger than H.

Remember that G has a normal series, namely the preimages of its central series. Write this series in ascending order, from 1 up to G.

Fix i so that N_i is the last subgroup wholly contained in H. Since H is a proper subgroup of G, there is another subgroup in the normal series, namely N_i+1, and N_i+1 contains elements outside of H.

Let y be an element in N_i+1-H, and let x represent a coset of N_i in H. By construction, cosets of N_i lie in the center of G/N_i iff they are contained in N_i+1. Since y represents such a coset, write yxN_i = xyN_i. Since N_i is normal, yN_i is the same as N_iy. Thus yxN_i = xN_iy. This holds for every x in H, hence y is an element that normalizes H, although y is not in H. The normalizer is properly larger than H.

Now let G be a finite nilpotent group and let J be a sylow subgroup. If J = G we are done. So let J be a proper subgroup of G. Let H be the normalizer of J in G. Recall that the normalizer of H is the same as H, which is a contradiction, unless H = G. Therefore every sylow subgroup is normal in G.

Next assume sylow subgroups are normal and let x be an arbitrary element of G. If x belongs to two separate sylow subgroups, its order is a power of a prime p, and a power of some other prime q, which is impossible. The various sylow subgroups are pairwise disjoint.

Let H and K be two sylow subgroups and let D be the group generated by H and K. Since H and K are disjoint and normal, D is the direct product of H and K. Since conjugation can be applied per component, D remains normal in G.

If J is a third sylow subgroup, it is disjoint from D, since everything in D has order divisible by p and q, and everything in J has order divisible by some other prime r. Again, D and J are both normal in G, and they generate a normal subgroup that is isomorphic to their direct product. Thus H J and K span a subgroup that is isomorphic to their 3-way direct product. This continues until G becomes the direct product of all its sylow subgroups.

Finally, assume G is the direct product of its sylow subgroups. Every finite p group is nilpotent, and the finite direct product of nilpotent groups is nilpotent, hence G is nilpotent.

When G is nilpotent, there is a generalization of the first sylow theorem. For every m dividing |G|, there is a subgroup of order m. If a sylow subgroup is based on the prime p, there is a subgroup of order p^k, where p^k divides m. This holds for all primes, and their direct product produces the desired subgroup of order m.

Let G be a finite group and let H be the intersection of its maximal subgroups. The inner automorphism x*G/x maps chains of subgroups to chains of subgroups. In other words, conjugation preserves the lattice of subgroups of G, and maps the intersection of maximal subgroups onto itself, hence H is normal in G.

Let K be another proper subgroup of G. What is K join H? Push K up to a maximal subgroup of G, and H join K can only increase. Yet H is contained in every maximal subgroup. Therefore any subgroup of H, joined with any proper subgroup of G, yields a proper subgroup of G.

Let P be a sylow subgroup of H. Remember that conjugation by x performs an automorphism on H. This moves P to a sylow subgroup of H that is isomorphic to P. Since all sylow subgroups are conjugate, some y in H carries the image of P back to P. Write this as yxP/x/y = P. In other words, yx is in the normalizer of P.

Let K be the normalizer of P in G, and note that H join K includes y inverse times yx. Since x was arbitrary, H join K is all of G. Yet this is a contradiction, unless K = G. Therefore the normalizer of P is G, and P is normal in G.

Restrict to H, and P is normal in H. This holds for all sylow subgroups of H, and by the previous theorem, H is nilpotent.

Hans Julius Zassenhaus proved a rather technical relationship among subgroups, which acts primarily as a lemma for the next theorem.

In what follows, the symbol ×, when applied to subgroups, means the join, or the subgroup generated by the two operands. The symbol ∩ denotes the intersection of two subgroups. You need to be familiar with the theorem that characterizes the join of two subgroups. It is an important part of this proof.

Instead of building a map one to the other, I will equate the first factor group with (A∩B) mod (A∩T) × (B∩S). This expression is symmetric. Swap A for B and S for T and find the very same group. Thus the reasoning below implies a parallel isomorphism from the second quotient onto the same group, and the two quotients are isomorphic to each other.

Start with S × (A∩B) mod S × (A∩T), and work towards our symmetric expression.

Since S is normal in A, restrict to B to show S∩B is normal in A∩B. Similarly, T∩A is normal in B∩A.

Let J be the join of the two groups S∩B and T∩A. Both groups are normal in A∩B, and these are the generators of J, hence J is also normal in A∩B. J is the kernel of the destination factor group.

With S normal in H, a group homomorphism maps H onto the cosets of S in H, which are faithfully represented by the cosets of S∩B in A∩B. Again, this was described in the theorem on joining subgroups.

Remember that J was defined as B∩S join A∩T. Thus J is a certain collection of cosets of B∩S, all of them living in A∩B. And J is a normal subgroup. So apply a second homomorphism that has J as its kernel. Call this composite homomorphism f. Now f maps H onto (A∩B)/J.

The kernel of f is J = (A∩T) × (B∩S), joined with S. Write this as (A∩T) × (B∩S) × S. Elements of these three subgroups generate the kernel. But the second is a subgroup of the third, i.e. contained in the third, so just write the kernel as (A∩T) × S. This is the kernel we want for H.

To recap, H mod S×(A∩T), which is the same as S×(A∩B) mod S×(A∩T), is isomorphic to (A∩B) mod (A∩T)×(B∩S). This is our destination factor group, symmetric in A and B, S and T. That completes the proof.

If you have to go over this three or four times to get it, you're not the only one!

This is the long awaited theorem that factors a finite group G uniquely into simple groups. It is often called the Jordan Holder theorem, because it was developed by Jordan and Holder.

Let U and V be two subnormal series for the group G. U₀ = V₀ = G, and U_k = V_l = 1.

Build a matrix M with the series U running down the left hand side, and the series V running across the top. Number the rows and columns of M starting at 0, consistent with U and V. Thus M has k+1 rows and l+1 columns.

Using the notation of the previous section, let M_i,j = U_i+1 × (U_i∩V_j). The bottom row is simply 1.

Watch what happens as you move across the i^th row of M. Move from column j to column j+1, and find the groups U_i+1×(U_i∩V_j) and U_i+1×(U_i∩V_j+1). Using the variables of the previous section gives S×(A∩B) and S×(A∩T). By Zassenhaus, the latter is normal in the former. The i^th row of M builds a subnormal series. The first entry in the i^th row is U_i+1×(U_i∩G), or U_i. The last entry is U_i+1×(U_i∩1), or U_i+1. Hence the i^th row of M is a subnormal series from U_i down to U_i+1.

Run the entire matrix from top to bottom, left to right. This produces a subnormal series from U₀ to U₁, from U₁ to U₂, from U₂ to U₃, and so on down to U_k, giving a subnormal series for all of G. This may not fit the technical definition, since many of the groups in the series are the same. But if you don't mind the duplicates, the groups form a descending chain from G all the way down to 1. The factor groups are U_i+1×(U_i∩V_j) mod U_i+1×(U_i∩V_j+1), for all i and j, save the last row and last column.

Next, build a parallel matrix N, with U and V as before, but this time let N_i,j = V_j+1×(V_j∩U_i). The rightmost column of N is 1. Walk down the j^th column of N and find a subnormal series from V_j down to V_j+1. Traverse the columns of N, from left to right, and build a subnormal series for all of G.

The factor groups in this series are V_j+1×(V_j∩U_i) mod V_j+1×(V_j∩U_i+1), for all i and j, save the last row and last column.

Now for the magic. Let i = 3 and j = 7, just to illustrate. The series from the matrix M produces the factor group U₄×(U₃∩V₇) mod U₄×(U₃∩V₈). The series from the matrix N produces the factor group V₈×(V₇∩U₃) mod V₈×(V₇∩U₄). By Zassenhaus, these quotient groups are the same. If M creates a nontrivial factor group at i,j, then N creates the same factor group at j,i, and conversely. Both matrices produce the same factor groups, albeit in a different order.

If U and V are composition series, there are no proper refinements. The refinements produced by M and N merely duplicate some of the subgroups in U, or the subgroups of V, respectively. Both M and N produce the same factor groups, many of them trivial. The nontrivial factors must agree, and these are the factor groups produced by U and by V.

If G has a composition series, the simple factor groups of G are unique up to order.

Every finite group G has a composition series, hence G factors uniquely into a set of simple groups.

A group G is uniserial if it has one and only one composition series. If G has one maximal normal subgroup, which has one maximal normal subgroup, and so on down to 1, then the factor groups are prescribed at each step. However, G could have many maximal normal subgroups. The abelian group (Z/2)ⁿ has a maximal normal subgroup for each space of dimension n-1, but no matter the chain of descending subgroups, each factor group is Z/2.

Uniserial groups include simple groups (by default), the finite direct product of a simple group (as above), and symmetric groups.