## Simple Modules, Jordan Holder

### Jordan Holder

There is a branch of mathematics devoted to the study of descending chains of subgroups. Each is a subgroup of the previous; in fact each is a normal subgroup of the previous. You don't need to know all the theorems in that section, but you should be familiar with chains of subgroups.

Each normal subgroup inside the previous group inplies a factor group, also called a quotient group. If the quotient groups are simple, and if the list is finite, The quotient groups form the decomposition series of G. For instance, Z30 contains Z15 contains Z3 contains 0 has the decomposition series Z2, Z5, Z3.

Like unique factorization into primes, G has a unique decomposition series. The factor groups may appear in a different order, depending on where you start, but you always get the same simple groups. This is called the Jordan Holder theorem.

We are going to prove the same thing for modules. After all, modules are groups with a ring action. In many ways the task has become easier, because all the groups are abelian. We don't have to ask whether Gi+1 is a normal subgroup of Gi, and whether they are both normal in G. Every submodule is "normal", and every submodule acts as the kernel of a module homomorphism, producing a quotient module. So all we need is a chain of submodules, and we're on our way.

Let's begin.

### Submodules and Quotient Modules

Let K be a submodule of M. The cosets of K in M form the quotient module, with K as kernel. Verify that the action of R on these cosets is well defined, giving a valid R module.

Like the correspondence theorems for groups, submodules in M/K correspond 1-1 with the submodules of M that contain K. Thus an infinite chain of submodules in M/K lifts to an infinite chain in M. And an infinite chain in K is an infinite chain in M. If M is noetherian or artinian, K and M/K are also noetherian or artinian.

### Refining the Chain

If M/K has a proper submodule V, let U be the preimage of V, whence M contains U contains K, and containment is proper. We have refined the chain by inserting an additional submodule. A chain can be refined iff one of its factors is not a simple module.

Let M be both noetherian and artinian, and suppose a chain of submodules can be refined again and again, forever. Assign 0 the number 0, and assign M the number 1, at the top. If you are given an initial chain of n submodules, give them the numbers (n-1)/n down to 1/n. Now start refining the chain. If an intermediate submodule is inserted between to modules having the numbers x and y, give the new module the number (x+y)/2. Thus each module has a rational number that describes its relative position in the chain. If U has a smaller number than V, U is a submodule of V. This is a linear ordering.

If the refinement is infinite, there are infinitely many values between 0 and 1. This implies a cluster point c between 0 and 1. Modules approach c from above, or below, or both. This contradicts the fact that M is noetherian and artinian.

Every decomposition series is finite, exhibiting simple modules as factors. Start with M and 0 and refine as far as possible to show that at least one decomposition series exists.

### Jordan Holder

The Zassenhaus lemma applies to all groups, finite or infinite, abelian or nonabelian. Applying it to modules is practically a corollary. We only need validate the action of R. The two quotient groups are isomorphic as groups; we need to show they are isomorphic as R modules.

Subgroups are now submodules, and hence, the homomorphisms that Zassenhaus built become R module homomorphisms. The destination group, (A∩B)/J, is a well-defined R module, and both quotients are isomorphic to this common module, hence they are isomorphic to each other.

Once Zassenhaus is extended to R modules, the Jordan Holder theorem follows, exactly as written. If it exists, the finite decomposition series for a module M is unique. And if M is noetherian and artinian, it always has a decomposition series.

Conversely, if M has a finite decomposition series (which must be unique up to order), it is noetherian and artinian. Suppose M has an infinite chain of submodules, ascending or descending, as you prefer. Let M1 be the first module in the decomposition series, with factor F1 = M/M1. Since F1 is simple, the infinite chain must be contained in M1. Perhaps the top of the descending chain = M1, but we still have an infinite chain inside M1. Repeat this reasoning for M2 inside M1, and M3 inside M2, and so on. Finally the last submodule contains an infinite chain. However, this module is simple, so we have a contradiction, and M is noetherian and artinian.

If M contains infinite chains, anything goes. Consider Z, as a Z module. Submodules could be 2Z, 4Z, 8Z, etc, giving factor groups of Z2 forever. Or submodules could be 3Z, 9Z, 27Z, etc, giving factor groups of Z3. There is no unique factorization here.