Krull Schmidt, Normal Endomorphisms

Normal Endomorphisms

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. Conjugate v(x) by a and get v(a*x/a), which is another point in the image of v. Therefore the image of v is normal in G.

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. See modules for a proof.

Fittings Decomposition Theorem

Let G be both acc and dcc and let f be a normal endomorphism. For some n, G is the direct product of the kernel, and the image, of fn.

Choose n so that kernels and images beyond n are constant. For notational convenience, let v = fn. 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 group identity). Kernel and image are disjoint.

For any c in G, v(c) = v2(d) for some d. Let b = c*v(1/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(y/x/y). 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.

Nilpotent

An endomorphism f is nilpotent if for some n, fn is trivial.

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 above. 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 ijective, and f becomes an automorphism, as described above.

The Ring of Endomorphisms

Let f1 f2 f3 … fk be a finite set of normal endomorphisms. Add two endomorphisms by letting (f1+f2)(x) = f1(x)*f2(x). 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 fk are independent projections. Naturally f1+f2 is a valid endomorphism in the world of modules.

Don't be confused by the plus sign; f1+f2 may 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. Also verify that the sum or product of normal endomorphisms is normal.

Assuming all sums of all products of f1 through fk create valid endomorphisms, we have built a ring of endomorphisms on the group G. This is similar to the ring of endomorphisms on a module M, although that is a true ring, whereas the ring of endomorphisms of G is a pseudo-ring, because addition may not be commutative.

Adding Nilpotent Endomorphisms

Assume G is indecomposable, and satisfies acc and dcc. Let the normal endomorphisms f1 through fk be nilpotent. If they build a ring of endomorphisms, as above, all the endomorphisms are nilpotent.

Consider the sum f1+f2. We showed above that this is either nilpotent or an automorphism. Suppose it is an automorphism, with an 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 endomorphism.

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

Write e1*(e1+e2) = e1 = (e1+e2)*e1, whence e1*e2 = e2*e1.

Expand (e1+e2)n. Addition may not be commutative, so you may have lots of terms. But multiplication is commutative, so each term is 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)n takes G into 1, 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 fk has a nontrivial kernel, and is nilpotent. By induction, every finite sum of these products, which is every function in our ring, is nilpotent.