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

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

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

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

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

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.) Because ${\displaystyle M}$  is indecomposable, one of those two summands must be equal to ${\displaystyle M}$  and the other must be the zero submodule. Depending on which of the two summands is zero, we find that ${\displaystyle f}$  is either bijective or nilpotent.^[2]

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