• Overview
• ACC, DCC
• Decomposable
• Endomorphisms
• Fittings Decomposition Theorem
• Nilpotent
• Ring of Endomorphisms
• Unique Decomposition
• Two Representations

The Jordan Holder theorem states that the composition series for a group is unique. If you start with a different kernel, and build a different composition series, you get the same factor groups, possibly in a different order.

Jordan Holder was generalized to modules without too much fuss. The quotient modules that arise from the composition series of M are uniquely a function of M. A different composition series will produce the same factors, possibly in a different order.

The Krull Schmidt theorem is similar. Write G as a direct product of smaller groups, and if the component groups are indecomposable, i.e. not a direct product of still smaller groups, then the composition is unique. There is only one way to write G as the finite direct product of component groups. For instance, Z/3 * Z/5 * Z/20 cannot be written as any other direct product.

Like Jordan Holder, Krull Schmidt generalizes to modules. A module M is the direct product of submodules in a unique way. As you move through the proof, take a moment to verify each step as it applies to modules. In many cases the algebra is easier, because the group suddenly becomes abelian. You only need turn group homomorphisms into module homomorphisms, and you're all set.

As a warmup, let M be a finitely generated module over a pid. As described in the previous chapter, M is the direct product of cyclic modules in a unique way. Let U be one of these cyclic modules. If U is part of the free submodule of M, then U is the same as R. If R is a direct product of two R modules, then R is the direct product of two rings. This is because R is commutative. Write 1 as e + f, according to the components of the direct product. Now 1*1 = e² + ef + fe + f² = e + f. When f acts on e the result lives in the first component, and when e acts on f the result lives in the second. This because each component is its own R submodule. But ef = fe, so the same element is in both components, and that means ef = fe = 0. Rewrite the equation as e² + f² = e + f. Within the first component, e² = e, and within the second component, f² = f. These are idempotents, that turn R into the direct product of two rings. However, R is an integral domain, and has no nontrivial idempotents. Therefore R is indecomposable.

Let U be one of the cyclic modules in M_T, isomorphic to R/p^j. Note that U is itself a ring, and U is a U module. The action of R is the same as the action of U. If U is a direct product of R modules then U is a direct product of U modules, and U is a direct product of rings. This again implies idempotents. Let x be an idempotent in U. If x is a multiple of p then x² cannot equal x. The exponent on p is doubled. So x is nonzero mod p. Since p is maximal, R/P is a field. If x² = x in a field, then x = 1. Write x as 1+yp. Square this and get 1 + 2yp + y²p². Look mod p², and y has to equal 2y. Thus y = 0. The same would happen with 1+yp², 1+yp³, etc. The only idempotent is 1, and U is indecomposable.

M has been separated into a unique finite direct product of indecomposable modules. Of course this is a special case, wherein R is a pid and each component is cyclic. In general, R is unconstrained, and a component can be just about anything you like. R could be noncommutative, whence M could be a left or right module. When applied to groups, G could be nonabelian. Still, there is only one way, up to isomorphism, to write a group or module as a finite direct product of indecomposable groups or modules.

Throughout this chapter, a "chain" consists of normal subgroups of G, linearly ordered by containment. This is different from the normal series that we saw earlier, where each subgroup is normal in the subgroup above it. The subgroups of this chain need only be normal in G.

If G has no infinite ascending chain it has the "ascending chain condition", also known as acc. If G has no infinite descending chain it has the "descending chain condition", also known as dcc. This is consistent with the corresponding definitions for modules.

Note that G could be acc and dcc, and still have infinite chains of subgroups that are not normal. Consider the alternating group on infinitely many letters. This group is simple, hence it is acc and dcc. The subgroups A_n, as n runs from 1 to infinity, form an ascending chain, and the subgroups that exclude the first n letters form a descending chain.

Let Q be a quotient group of G. Since normal subgroups of Q pull back to normal subgroups of G, an infinite chain in Q lifts to an infinite chain in G. Therefore Q inherits acc or dcc from G.

Let G = K*Q, a special case of kernel and quotient. By the above, Q and K inherit acc or dcc from G. The converse is also true. If K and Q are acc or dcc then so is G. See the corresponding proof for modules.

This generalizes to a finite direct product of groups. The product is acc or dcc iff the same holds for each component.

Again, let G = K*Q. Let J be a normal subgroup of K. Conjugate x*J/x, and the K component maps J onto J, while the Q component drops out. Thus J is normal in G. A chain of normal subgroups in K becomes, without modification, a chain in G.

A group is decomposable if it is the direct product of two nontrivial groups. Simple groups are indecomposable, but so is Z/pⁿ and S_n. By convention an indecomposable group is nontrivial, just as the prime numbers do not include 1.

A group is finitely decomposable if it is the finite direct product of indecomposable groups. Thus an indecomposable group is finitely decomposable, even though it is not decomposable.

A finite direct product of finitely decomposable groups is finitely decomposable.

Assume G is not finitely decomposable. This means it is not indecomposable, hence it is decomposable. Separate it into H*K. One of these components, H or K, is not finitely decomposable, so separate that into two pieces. One of those components can then be separated, and so on forever. This builds a descending chain of normal subgroups of G. The complement is an ascending chain. Either acc or dcc forces G to be finitely decomposable.

An endomorphism f is normal if it commutes with every inner automorphism. That is, f(a*x/a) = a*f(x)/a.

If you're generalizing this theorem to modules, * becomes +, the group is abelian, and every endomorphism is normal. So you can cross out the word normal and apply this theorem, and subsequent theorems, to all module endomorphisms. But for now, let's return to groups.

Let v be the composite endomorphism fⁿ. Note that v is a normal endomorphism.

Let H be normal in G, and conjugate f(H). That's the same as f applied to the conjugate of H, or f(H). Thus the image of a normal subgroup is normal.

Turn this around and let H be the preimage of J, with J normal in G. Suppose some x in H has a*x/a outside of H. Apply f and get y outside of J. Thus a*f(x)/a = y, which contradicts J normal in G. The preimage of a normal subgroup is normal.

Start with G and apply f repeatedly, and build a descending chain. Or start with e and take preimages repeatedly, and build an ascending chain. Since G and e are both normal in G, each group in the chain is indeed normal in G, as it should be.

If f is a monomorphism and G is dcc, f is an automorphism. Similarly, if f is an epimorphism and G is acc, f is an automorphism. The proof for modules applies here.

Wait a minute. Don't the kernels get larger as the images get smaller? Isn't acc or dcc enough? Well yes, if G is finite, or if the image of f is finite. Then ascending kernels correspond to descending images. But in general this doesn't work. Let M be an infinite dimensional vector space, with a basis b₀ b₁ b₂ b₃ etc. Let f push these basis elements forward, f(b₀) = b₁, f(b₁) = b₂, and so on. Invoke f again and again, and the kernel is always 0, while the images descend. Conversely, let f pull the basis back, with f(b₁) = b₀, and f(b₀) = 0. The image is always M, but the kernels ascend. So we need G to be both acc and dcc. Then there is some v() where the kernel never increases and the image never decreases.

Suppose y lies in both the kernel of v and the image of v. Let v(x) = y, so that x is in the kernel of v². Yet this kernel is the same as the kernel of v, hence y = 1 (the identity element for G). The kernel and the image are disjoint.

For any c in G, v(c) = v²(d) for some d. This because v and v² have the same image. Let b = c/v(d). Expand v(b) into v(c)/v²(d), which is 1. Thus b is in the kernel of v. Since b*v(d) = c, our arbitrary element c is something in the kernel times something in the image.

Let x lie in the kernel and let y lie in the image, and consider the commutator b = xy/x/y. Apply v and get v(x) * v(y/x/y). Of course v(x) drops out. Since v is normal this is the same as y*v(1/x)/y. This collapses to 1, hence b is in the kernel.

Remember that the image is normal in G, hence xy/x is in the image, and y is in the image, so b is in the image. With b in kernel and image, b = 1, and x and y commute. That completes the proof.

An endomorphism f is nilpotent if for some n, fⁿ maps all of G to 1. This is consistent with a nilpotent transformation on a vector space.

If G is indecomposable, and acc and dcc, and f is a normal endomorphism, then f is an automorphism or it is nilpotent. Let v = fⁿ, and let G = the kernel of v cross the image of v, as described in the previous section. Since G is indecomposable, either the kernel or the image is 1. If the image is 1 then f is nilpotent. If the kernel is 1 then f is injective, and f becomes an automorphism, as described earlier.

Let f₁ f₂ f₃ … f_k be a finite set of normal endomorphisms. Add two endomorphisms together by letting (f₁+f₂)(x) = f₁(x)*f₂(x). (The use of + is a bit confusing here, since + might not be commutative.) Assume the sum of any distinct subset of these endomorphisms, in any order, gives a well defined endomorphism. This does not always hold, but it is valid in certain situations, such as when G is abelian, or when f₁ through f_k are independent projections. Naturally f₁+f₂ is a valid endomorphism in the world of modules.

If f₁ + f₂ is a group homomorphism, we have the following for any x and y in G.

f₁(x)*f₁(y) * f₂(x)*f₂(y) = f₁(x)*f₂(x) * f₁(y)*f₂(y)

Cancel f₁(x) from the left and f₂(y) from the right and get this.

f₁(y) * f₂(x) = f₂(x) * f₁(y)

Set y = x, and f₁ + f₂ = f₂ + f₁. Thus addition is commutative after all, and we are justified in using the + operator. Of course, if f₁ + f₂ is not a group homomorphism, just a function from G into G, then f₁ + f₂ might not equal f₂ + f₁.

In the algebra of endomorphisms, let f₁+f₂ denote the sum described above, and let f₁*f₂ denote f₁ followed by f₂. Verify that multiplication distributes over addition, on either side. Thus the following expansion is valid.

(f₁ + f₂) (f₃ + f₄) = f₁f₃ + f₂f₃ + f₁f₄ + f₂f₄

Verify that the sum or composition of normal endomorphisms is normal (assuming the sum is an endomorphism). We already addressed function composition, but here it is again in detail.

f₁(f₂(a*x/a)) = f₁(a*f₂(x)/a) = a*f₁(f₂(x))/a

f₁(a*x/a) * f₂(a*x/a) = a*f₁(x)/a * a*f₂(x)/a = a*f₁(x)*f₂(x)/a

Assume G is indecomposable, and satisfies acc and dcc. Let the normal endomorphisms f₁ through f_k be nilpotent. All the derived endomorphisms are also nilpotent.

We showed above that any normal endomorphism is either nilpotent or an automorphism. If f₁ is nilpotent it is not an automorphism, and it has a kernel. Thus f₁f₂ has a kernel, and is a normal endomorphism, and is nilpotent. All finite function compositions are nilpotent.

Suppose the sum f₁ + f₂ is an automorphism,with inverse e. If d = f₁ + f₂, we have de = ed = 1. Remember that d is normal. Let c be an inner automorphism, whence cd = dc. Multiply by e on the left and right, whence ec = ce. Therefore e is a normal automorphism.

Let e₁ = e*f₁, and e₂ = e*f₂. Thus e₁+e₂ is the identity automorphism.

By composition, e₁ and e₂ are normal.

Write e₁(e₁+e₂) = e₁ = (e₁+e₂)e₁, thus e₁e₂ = e₂e₁.

Expand (e₁+e₂)ⁿ. Since e₁ and e₂ commute, each product becomes a power of e₁ times a power of e₂. Since f₁ and f₂ are nilpotent, e₁ and e₂ have nontrivial kernels. They cannot be automorphisms, hence they are nilpotent. Choose n large enough to make e₁ⁿ and e₂ⁿ trivial, and double it. Now (e₁+e₂)²ⁿ takes G into 1, term by term, whence e₁+e₂ is nilpotent. This contradicts the fact that e₁ + e₂ is the identity automorphism. Therefore f₁ + f₂ is nilpotent.

Any product of functions drawn from f₁ through f_k is nilpotent, and any finite sum of these products that is a valid group endomorphism is nilpotent.

Let G be nontrivial, satisfying acc and dcc. Being acc and dcc, there is at least one finite decomposition G = G₁*G₂*G₃*…G_k. Suppose G is isomorphic to another direct product H₁*H₂*H₃*…H_l, where all component groups are indecomposable. Restrict the isomorphism to H_i, giving an isomorphic subgroup inside G. Do this across the board, and say, without loss of generality, that G = H₁ * H₂ * H₃ * … H_l. In other words, G itself has two decompositions, without vectoring through some other group that is isomorphic to G. We will prove that k = l, and after reindexing, G_i and H_i are isomorphic.

Why not just say G_i = H_i? Well, as an example, G could be Z*Z. Let G₁ and G₂ be the components of G. Then let [1,2] generate H₁, and let [1,1] generate H₂. Corresponding subgroups are isomorphic, but not equal. So G decomposes into a unique direct product, up to isomorphism.

Let w₀ be the assertion that G is the product of H₁ through H_l, which is true by assumption. For j between 1 and l, let w_j state that, after reindexing, G is the product of G₁ through G_j times H_j+1through H_l, such that G_i and H_i are isomorphic up to j. This reindexing is backward compatible. Once the connections are made through j, they are left alone, as we find a new H_i in G that is isomorphic to G_j+1.

Assume w_j-1 is true. Let p_i project G onto G_i using the original decomposition. Let q_i project G onto G_i for i < j, and onto H_i for i ≥ j. In other words, q reflects the decomposition of w_j-1.

View p_i and q_i as endomorphisms of G. All endomorphisms are projections, and normal.

The product of any two projections from the set p, or the set q, is trivial for distinct projections, or it is the projection when that projection is multiplied by itself.

Add two distinct projections and get a combined projection that extracts both components, such as G₃ and G₅. These are well defined endomorphisms. This builds a ring of endomorphisms, as described in the previous section.

Compose the identity automorphism, the sum over q_i, with p_j. This gives p_j = the sum of q_ip_j, as i runs from 1 to l. But when i < j, q_i =p_i, and p_ip_j is trivial; so let the sum run from j to l. The terms of the sum are normal endomorphisms, and they can be added together (as projections followed by p_j) to create well defined endomorphisms.

All these endomorphisms map G into G_j. The same is true if you restrict the domain to G_j. This gives endomorphisms on an indecomposable module, which happens to be acc and dcc. The results of the previous section are in force.

The sum of the component endomorphisms is nilpotent if each endomorphism is nilpotent. Yet p_j fixes G_j, and is an automorphism, hence not nilpotent. For some i, v = q_ip_j is not nilpotent. It is an automorphism on G_j.

Turn this around and let u = p_jq_i, an endomorphism on H_i. If u is nilpotent with order n, then vⁿ⁺¹ = q_iuⁿp_j. With uⁿ = 1, vⁿ⁺¹ = 1, which contradicts the fact that v is an automorphism; therefore u is also an automorphism. Both functions q_i and p_j must be injective and surjective, hence they build isomorphisms between G_j and H_i.

Reindex H, so that H_i becomes H_j.

For illustration, I will let j = 5 and l = 9. We still have to replace H₅ with G₅, and find a direct product that builds G.

Suppose a sum, a_i drawn from G_i, as i runs from 1 to 5, plus b_i drawn from H_i, as i runs from 6 to 9, equals 0. Apply q₅ to this equation, which is one of 9 projections in the set q. Everything drops out except for q₅(a₅). Remember that a₅ lives in G₅, and q₅ is an isomorphism from G₅ onto H₅. This gives the equation q₅(a₅) = 0. Since a₅ is nonzero and q₅ is an isomorphism, this is a contradiction. That takes a₅ out of the picture. The remaining subgroups are disjoint, as per w_j-1. Therefore the subgroups are all disjoint.

Next show that all of G is spanned. By assumption, G₁ through G₄ and H₅ through H₉ span G. The projection q₅ maps G₅ onto H₅. Start with G₅ and we have H₅ plus maybe some other stuff that is spanned by H₆ through H₉. Subtract that stuff away and find all of H₅, hence G is spanned.

Finally G₅ must commute with the other groups. It already commutes with G₁ through G₄. Suppose it does not commute with H₆ or higher. Let a in G₅ be b+y, where b is in H₅ and y is in the span of H₆ through H₉. Select z in H₆ through H₉ so that z and y do not commute. (We already know that z and b commute.) Slice z up according to its projections into G₁ through G₇. Only the slice belonging to G₅ might not commute with a in G₅. Call this slice z₅, and ignore the other projections. Apply q₅ to see what part of z₅ lies in H₅. Of course no part of z lies in H₅, hence q₅(z₅) = 0. Yet q₅ is an isomorphism, hence z₅ = 0. Therefore z commutes with b+y, all subgroups commute, G is the direct product of G₁ through G₅ times H₆ through H₉, and w₅ is true.

Continue this process until j = k or l, whichever is less. At this point G = G times some other groups, and those other groups can only be trivial, thus k = l. The representation of G as a direct product of indecomposable groups is unique.

As a bonus, replacing H_j with G_j is an automorphism on G. Both H_j and G_j commute with all the G subgroups before, and all the H subgroups after, and q_j builds an isomorphism from G_j onto H_j, hence this isomorphism, along with the identity map on all the other subgroups, becomes an automorphism on G.

When G is a module, this theorem can assert the uniqueness of certain indecomposable summands, even if G is not acc or dcc. Assume G has an indecomposable summand G₁ that happens to be acc and dcc, even if G is not. This is the first in a finite decomposition G_i, while H_i is some other finite decomposition for G. The above proof is valid, at least for G₁, and builds an automorphism from G₁ through H₁ and back onto G₁. This path is injective from start to finish. Let F be the image of G₁ in H₁, and let E be the kernel of p₁(H₁) onto G₁. Since F is a perfect copy of G₁ inside H₁, the map p₁ is a semidirect product. Since H₁ is a module, the semidirect product becomes a direct product. However, H₁ is indecomposable, hence E = 0, and F is all of H₁. The map is once again an isomorphism, and replacing H₁ with G₁ builds an automorphism on G. The quotient modules G/G₁ and G/H₁ are isomorphic.

This extends to all the summands in the first list that are acc and dcc. G₁, G₂, and G₃ correspond to H₁, H₂, and H₃, and the resulting quotient modules are isomorphic.

If there are only 4 modules in the list, so that G₄ is G mod G₁*G₂*G₃, then replace the first 3 modules, build the associated isomorphism, and equate quotient modules, whence G₄ and H₄ are isomorphic. The decomposition is unique if all but one of the summands in either list is acc and dcc.

This chapter would not be complete without an example of a group that splits in two different ways. Obviously the group is not both acc and dcc.

Represent the group as Z*G, where Z is an abelian group and G is a nonabelian group. Eventually Z will turn into Z, the integers. I will use additive notation for Z and multiplicative notation for G. Thus 0 is the identity in Z and e is the identity in G. An ordered pair will be written [a,x], for a in Z and x in G. The bottom of the group is [0,G]. The goal is to rewrite this group as another direct product H*K.

A homomorphism carries H into Z, with kernel H₀ mapping to 0. In other words, H₀ = H intersect [0,G]. Similarly, K₀ = K intersect [0,G]. Thus H₀ and K₀ are subgroups of G that commute. In fact their direct product H₀ * K₀ is a subgroup of G. Call this subgroup S.

Let H_a be the coset of H₀ in G that corresponds to a in Z. This is another slice of H. It is H₀*x_a, where x_a is the cosrep. Or it is x_a*H₀. It is the same coset; x_a conjugates the kernel H₀ onto itself. There is such a coset for those elements a in Z that are projected from H. The same coset could be used again and again, the elements of Z make them distinct in Z*G. In fact every cosrep could be e, but that makes H the direct product of H₀ and a subgroup of Z, and hence decomposable. If there is more to H than [0,H₀], then some of the cosreps are outside of H₀.

The cosets are compatible with the action of Z. The order of x_a is the least power of x_a that winds up back in H₀. This could be 1, if x_a = e. The order of x_a divides the order of a in Z. If n×a = 0, then x_aⁿ corresponds to 0 in Z, and has to lie in H₀ by the definition of H₀.

K is K₀ times y_b for various values of b in Z.

If H₀ = G, every cosrep may as well be e, and H is G cross a subgroup of Z. This subgroup has to be 0, else H is a direct product. Thus H = [0,G]. That leaves K₀ = e, and if K includes two values of G for some b in Z, then K₀ is nontrivial, so K has one value of G corresponding to each b in Z. K is Z cross some individual members of G whose action is compatible with that of Z. The simplest example is Z cross e. But if Z is in fact the integers, 1 could correspond to any free element of G. In any case, K is isomorphic to Z, so we still have Z cross G up to isomorphism. Therefore H₀ and K₀ are nontrivial proper subgroups of G.

Represent H * K, projected into G, as H₀x_a * K₀y_b over every a and b. If all the cosets of H₀, and all the cosets of K₀, lie in S, then S is all we get from H * K. S = G, and G is a direct product H₀*K₀. So there are some cosreps outside of S.

A cosrep x_a conjugates H₀ onto itself, and commutes with K₀, thus conjugates S onto itself. The same for y_b, hence S is normal in G. The quotient group G/S is defined by the cosreps x_ay_b.

Both X_a and H₀ commute with K₀, and since H and K commute, x_a also commutes with y_b.

Suppose distinct cosets represented by x, cross the distinct cosets represented by y, produce distinct cosets of S in G. This makes G the direct product of H and K. Since G is indecomposable, some x_ay_b = x_cy_d, for x_a ≠ x_c and y_b ≠ y_d. Bring x to the left and y to the right, and there is some overlap in x and y, besides x₀ = y₀ = e. At first this looks like an intersection between H and K, but if a ≠ b, this is not the case. The components in Z keep them apart.

Note that the overlap could be S itself. That is, x_a = y_b = e, for some a ≠ b (to keep H and K apart).

Suppose every y_b overlaps with some x_a. The cosreps x_a are sufficient to cover all the cosets of S in G. You don't need y_b at all. G becomes the direct product of {H₀*x_a} and K₀. Thus there are some cosets from either side that do not overlap.

If Z is finitely generated then the quotient group G/S is finitely generated, by x_a and y_b, where a and b range over the generators of Z. Since x and y commute, the group is abelian, and G/S is a finite direct product of cycles. If any one of these cycles commutes with S then G splits. So it is important that some x not commute with H₀, and some y not commute with K₀.

Let Z be the integers. For notational convenience, let x = x₁ and let y = y₁. Both x and y are outside of S, and x and y are not the same coset of S, else all of y overlaps with x.

H includes [1,x], but H*K has to include [0,x]. There is some power p (possibly negative) such that (x/y)^p = x. Thus x^p-1 = y^p. We knew there had to be some overlap between x and y; here it is. If p = 1 then y is in S. If p = 2 then x and all powers of x are subsumed by y. Similarly p cannot equal 0 or -1. Try p = 3, so that x² = y³.

Now it's time to build G H and K, and see if it plays out. Represent H₀ and K₀ by two independent triangles. H₀ rotates its triangle, a group of order 3. K₀ flips (reflects) its triangle through the altitude, a group of order 2. Thus S is a group of order 6, H₀ * K₀.

As x passes through S, from xS to Sx, flip the first triangle and preserve the second. Thus x conjugates H₀ and commutes with K₀ as required. Note that x²S = Sx².

Pass y through S, from yS to Sy, and rotate the second triangle, while preserving the first. Note that y³S = Sy³.

Finally replace y³ with x² wherever it appears. This builds the group G, with words in canonical format: H₀ * K₀ times a power of x times a power of y. The exponent on x is any integer; the exponent on y is 0 1 or 2. The inverse of y is x^-2y² The quotient G/S is generated by y²/x. If G splits then G/S splits, but a group isomorphic to Z cannot split. So some part of the kernel would have to split off. Yet nothing in the kernel commutes with both x and y. Therefore G is indecomposable. Since Z is also indecomposable, we have our first representation.

Build H from H₀ and the powers of x, where the exponent on x corresponds to the integer in Z. Build K from K₀ and the powers of y, where the exponent on y corresponds to the integer in Z, but then replace each y³ with x². Thus x⁸y² corresponds to 14. These are valid subgroups of Z cross G.

Since H₀ and K₀ commute, and x and y commute, H and K commute.

H and K are disjoint. Suppose they have an element in common, with projection a in Z. a has to be a multiple of 3, else the element from K contains y. If a = 3n, then H presents x³ⁿ while K presents x²ⁿ. Therefore n = a = 0. This pulls H down to H₀ and K down to K₀, which are disjoint.

H and K span. Multiply [3,x³] by [-3,x^-2] to get [0,x] in [0,G]. Multiply [1,y] by [-1,x] to get y/x in G. The presence of y/x and x implies y. Thus [0,G] is covered. All the integers of Z are covered, and the group is spanned.

H has a quotient generated by x, which is cyclic, and does not split. And nothing in H₀ commutes with x, hence H is indecomposable. The same goes for K, with quotient generated by y.

Neither H nor K is abelian, hence neither can be isomorphic to Z. This gives the second, nonisomorphic representation for the same group.

In this example every group has a finite kernel and a free quotient, hence acc, but not dcc.