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

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

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

We claim that ${\displaystyle \mathrm {ker} \left(f^{k}\right)\cap \mathrm {im} \left(f^{k}\right)=0}$ . Indeed, 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 ${\displaystyle x=f^{k}\left(y\right)=0}$ .

Moreover, ${\displaystyle \mathrm {ker} \left(f^{k}\right)+\mathrm {im} \left(f^{k}\right)=M}$ : 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 ${\displaystyle f^{k}\left(x-f^{k}\left(y\right)\right)=f^{k}\left(x\right)-f^{2k}\left(y\right)=0}$ , 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, ${\displaystyle M}$  is the direct sum of ${\displaystyle \mathrm {im} (f^{k})}$  and ${\displaystyle \mathrm {ker} (f^{k})}$ . (This statement is also known as the Fitting decomposition theorem.^[2]) Because ${\displaystyle M}$  is indecomposable, one of those two summands must be equal to ${\displaystyle M}$ , and the other must be the trivial submodule. Depending on which of the two summands is zero, we find that ${\displaystyle f}$  is either bijective or nilpotent.^[3]

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