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			The '''Fitting lemma''', named after the mathematician [[Hans Fitting]], is a basic statement in [[abstract algebra]]. Suppose ''M'' is a [[module (mathematics)\|module]] over some [[ring (mathematics)\|ring]]. If ''M'' is [[indecomposable module\|indecomposable]] and has finite [[length of a module\|length]], then every [[endomorphism]] of ''M'' is either [[bijective]] or [[nilpotent]].

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			A version of Fitting's lemma is often used in the [[group representation\|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''.

			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''<sup> 2</sup>), im(''f''<sup> 3</sup>),..., the second sequence is the ascending sequence ker(''f''), ker(''f''<sup> 2</sup>), ker(''f''<sup> 3</sup>),.... Because ''M'' has finite length, the first sequence cannot be ''strictly'' decreasing forever, so there exists some ''n'' with im(''f''<sup> ''n''</sup>) = im(''f''<sup> ''n''+1</sup>). Likewise (as ''M'' has finite length) the second sequence cannot be ''strictly'' increasing forever, so there exists some ''m'' with ker(''f''<sup> ''m''</sup>) = ker(''f''<sup> ''m''+1</sup>). It is easily seen that im(''f''<sup> ''n''</sup>) = im(''f''<sup> ''n''+1</sup>) yields im(''f''<sup> ''n''</sup>) = im(''f''<sup> ''n''+1</sup>) = im(''f''<sup> ''n''+2</sup>) = ..., and that ker(''f''<sup> ''m''</sup>) = ker(''f''<sup> ''m''+1</sup>) yields ker(''f''<sup> ''m''</sup>) = ker(''f''<sup> ''m''+1</sup>) = ker(''f''<sup> ''m''+2</sup>) = ... . Putting ''k'' = max(''m'',''n'' ), it now follows that im(''f''<sup> ''k''</sup>) = im(''f''<sup> 2''k''</sup>) and ker(''f''<sup> ''k''</sup>) = ker(''f''<sup> 2''k''</sup>). Hence, <math>\mathrm{ker}\left(f^k\right) \cap \mathrm{im}\left(f^k\right) = 0</math> (because every <math>x\in \mathrm{ker}\left(f^k\right) \cap \mathrm{im}\left(f^k\right)</math> satisfies <math>x=f^k\left(y\right)</math> for some <math>y\in M</math> but also <math>f^k\left(x\right)=0</math>, so that <math>0=f^k\left(x\right)=f^k\left(f^k\left(y\right)\right)=f^{2k}\left(y\right)</math>, therefore <math>y\in\mathrm{ker}\left(f^{2k}\right)=\mathrm{ker}\left(f^k\right)</math> and thus <math>0=f^k\left(y\right)=x</math>) and <math>\mathrm{ker}\left(f^k\right) + \mathrm{im}\left(f^k\right) = M</math> (since for every <math>x\in M</math>, there exists some <math>y\in M</math> such that <math>f^k\left(x\right)=f^{2k}\left(y\right)</math> (since <math>f^k\left(x\right)\in\mathrm{im}\left(f^k\right)=\mathrm{im}\left(f^{2k}\right)</math>), and thus <math>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</math>, so that <math>x-f^k\left(y\right)\in\mathrm{ker}\left(f^k\right)</math> and thus <math>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)</math>). Consequently, ''M'' is the [[direct sum of modules\|direct sum]] of im(''f''<sup> ''k''</sup>) and ker(''f''<sup> ''k''</sup>). 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.<ref>Jacobson (2009), p. 113–114.</ref>

			== Notes ==
			<references/>

			== References ==
			* {{Citation\| last=Jacobson\| first=Nathan\| author-link=Nathan Jacobson\| year=2009\| title=Basic algebra\| edition=2nd\| volume = 2 \| series= \| publisher=Dover\| isbn = 978-0-486-47187-7}}

			{{DEFAULTSORT:Fitting Lemma}}
			[[Category:Module theory]]
			[[Category:Lemmas]]

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