# Fitting lemma

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.[1]

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 im(f), im(f 2), im(f 3),…,
• the second sequence is the ascending sequence ker(f), ker(f 2), ker(f 3),…

Because M has finite length, the first sequence cannot be strictly decreasing forever, so there exists some n with im(f n) = im(f n+1). Likewise (as M has finite length) the second sequence cannot be strictly increasing forever, so there exists some m with ker(f m) = ker(f m+1). It is easily seen that im(f n) = im(f n+1) yields im(f n) = im(f n+1) = im(f n+2) = …, and that ker(f m) = ker(f m+1) yields ker(f m) = ker(f m+1) = ker(f m+2) = …. Putting k = max(m,n), it now follows that im(f k) = im(f 2k) and ker(f k) = ker(f 2k). Hence, ${\displaystyle \mathrm {ker} \left(f^{k}\right)\cap \mathrm {im} \left(f^{k}\right)=0}$  (because every ${\displaystyle x\in \mathrm {ker} \left(f^{k}\right)\cap \mathrm {im} \left(f^{k}\right)}$  satisfies ${\displaystyle x=f^{k}\left(y\right)}$  for some ${\displaystyle y\in M}$  but also ${\displaystyle f^{k}\left(x\right)=0}$ , so that ${\displaystyle 0=f^{k}\left(x\right)=f^{k}\left(f^{k}\left(y\right)\right)=f^{2k}\left(y\right)}$ , therefore ${\displaystyle y\in \mathrm {ker} \left(f^{2k}\right)=\mathrm {ker} \left(f^{k}\right)}$  and thus ${\displaystyle 0=f^{k}\left(y\right)=x}$ ) and ${\displaystyle \mathrm {ker} \left(f^{k}\right)+\mathrm {im} \left(f^{k}\right)=M}$  (since for every ${\displaystyle x\in M}$ , there exists some ${\displaystyle y\in M}$  such that ${\displaystyle f^{k}\left(x\right)=f^{2k}\left(y\right)}$  (since ${\displaystyle f^{k}\left(x\right)\in \mathrm {im} \left(f^{k}\right)=\mathrm {im} \left(f^{2k}\right)}$ ), and thus ${\displaystyle f^{k}\left(x-f^{k}\left(y\right)\right)=f^{k}\left(x\right)-f^{k}\left(f^{k}\left(y\right)\right)=f^{k}\left(x\right)-f^{2k}\left(y\right)=0}$ , so that ${\displaystyle x-f^{k}\left(y\right)\in \mathrm {ker} \left(f^{k}\right)}$  and thus ${\displaystyle x\in \mathrm {ker} \left(f^{k}\right)+f^{k}\left(y\right)\subseteq \mathrm {ker} \left(f^{k}\right)+\mathrm {im} \left(f^{k}\right)}$ ). Consequently, M is the direct sum of im(f k) and ker(f k). Because M is indecomposable, one of those two summands must be equal to M, and the other must be equal to {0}. Depending on which of the two summands is zero, we find that f is bijective or nilpotent.[2]

## Notes

1. ^ Jacobson, A lemma before Theorem 3.7.
2. ^ Jacobson (2009), p. 113–114.

## References

• Jacobson, Nathan (2009), Basic algebra, 2 (2nd ed.), Dover, ISBN 978-0-486-47187-7