BREAKING NEWS

## Summary

The Fitting lemma, named after the mathematician Hans Fitting, is a basic statement in abstract algebra. Suppose M is a module over some ring. If M is indecomposable and has finite length, then every endomorphism of M is either an automorphism or nilpotent.

As an immediate consequence, we see that the endomorphism ring of every finite-length indecomposable module is local.

A version of Fitting's lemma is often used in the representation theory of groups. This is in fact a special case of the version above, since every K-linear representation of a group G can be viewed as a module over the group algebra KG.

## Proof

To prove Fitting's lemma, we take an endomorphism f of M and consider the following two sequences of submodules:

• The first sequence is the descending sequence $\mathrm {im} (f)\supseteq \mathrm {im} (f^{2})\supseteq \mathrm {im} (f^{3})\ldots$ ,
• the second sequence is the ascending sequence $\mathrm {ker} (f)\subseteq \mathrm {ker} (f^{2})\subseteq \mathrm {ker} (f^{3})\ldots$

Because $M$  has finite length, both of these sequences must eventually stabilize, so there is some $n$  with $\mathrm {im} (f^{n})=\mathrm {im} (f^{n^{\prime }})$  for all $n^{\prime }\geq n$ , and some $m$  with $\mathrm {ker} (f^{m})=\mathrm {ker} (f^{m^{\prime }})$  for all $m^{\prime }\geq m$ .

Let now $k=\max\{n,m\}$ , and note that by construction $\mathrm {im} (f^{2k})=\mathrm {im} (f^{k})$  and $\mathrm {ker} (f^{2k})=\mathrm {ker} (f^{k})$ .

We claim that $\mathrm {ker} \left(f^{k}\right)\cap \mathrm {im} \left(f^{k}\right)=0$ . Indeed, every $x\in \mathrm {ker} \left(f^{k}\right)\cap \mathrm {im} \left(f^{k}\right)$  satisfies $x=f^{k}\left(y\right)$  for some $y\in M$  but also $f^{k}\left(x\right)=0$ , so that $0=f^{k}\left(x\right)=f^{k}\left(f^{k}\left(y\right)\right)=f^{2k}\left(y\right)$ , therefore $y\in \mathrm {ker} \left(f^{2k}\right)=\mathrm {ker} \left(f^{k}\right)$  and thus $x=f^{k}\left(y\right)=0$ .

Moreover, $\mathrm {ker} \left(f^{k}\right)+\mathrm {im} \left(f^{k}\right)=M$ : for every $x\in M$ , there exists some $y\in M$  such that $f^{k}\left(x\right)=f^{2k}\left(y\right)$  (since $f^{k}\left(x\right)\in \mathrm {im} \left(f^{k}\right)=\mathrm {im} \left(f^{2k}\right)$ ), and thus $f^{k}\left(x-f^{k}\left(y\right)\right)=f^{k}\left(x\right)-f^{2k}\left(y\right)=0$ , so that $x-f^{k}\left(y\right)\in \mathrm {ker} \left(f^{k}\right)$  and thus $x\in \mathrm {ker} \left(f^{k}\right)+f^{k}\left(y\right)\subseteq \mathrm {ker} \left(f^{k}\right)+\mathrm {im} \left(f^{k}\right)$ .

Consequently, $M$  is the direct sum of $\mathrm {im} (f^{k})$  and $\mathrm {ker} (f^{k})$ . (This statement is also known as the Fitting decomposition theorem.) Because $M$  is indecomposable, one of those two summands must be equal to $M$ , and the other must be the trivial submodule. Depending on which of the two summands is zero, we find that $f$  is either bijective or nilpotent.