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The theorem of Maschke tells us that every representation of a finite group is the direct sum of irreducible representation. More precisely:

Let $G$ be a finite group, $K$ a field whose characteristic does not divide the order of $G$, $V$ a $K$-vector space of finite dimension and $\rho:G\to {\rm GL}(V)$ a representation of $G$. If $W$ is a $G$-invariant vector space of $V$, then there exists a $G$-invariant complement $W'$ such that $V=W\oplus W'$.

Now, I would like to prove or to find a counter-example to the following generalisation:

Let $G$ be a group such that every element has finite order, $K$ a field of characteristic $0$, $V$ a $K$-vector space of finite dimension and $\rho:G\to {\rm GL}(V)$ a representation of $G$. If $W$ is a $G$-invariant vector space of $V$, then there exists a $G$-invariant complement $W'$ such that $V=W\oplus W'$.

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  • $\begingroup$ @The User: why do they generate a finite subgroup? $\endgroup$
    – Vladimir Dotsenko
    Commented May 14, 2013 at 10:17
  • $\begingroup$ If the finite orders of elements are uniformly bounded, this was proved by Burnside. It seems a bit too much to believe that it would work otherwise. I would speculate that one should be construct a counter-example using a construction of Golod-Shafarevich of such groups from nil-algebras... $\endgroup$
    – Vladimir Dotsenko
    Commented May 14, 2013 at 10:19
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    $\begingroup$ This was exactly not proved by Burnside. The assertion that a finitely generated group with $x^n=1$ for all elements x is finite, is exactly Burnside's problem and is known to be false for large enough exponents. However: This might just work out if we do not work in $G$ itself, but in the image of $G$ in $GL(V)$, because linear groups are much better behaved than general infinite groups when it comes to these kinds of problems. I do not know off the top of my head whether or not Burnside's problem is true for linear groups or not. $\endgroup$
    – Johannes Hahn
    Commented May 14, 2013 at 12:15
  • $\begingroup$ @Vladimir You are right, in this general setting we cannot assume that. $\endgroup$
    – The User
    Commented May 14, 2013 at 12:19
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    $\begingroup$ @Johannes Hahn: I was referring to the result on linear groups which was indeed proved by Burnside himself: a linear group with $x^n=1$ for all $x$ is finite. (It is weaker than the result of Schur on periodic groups mentioned below which is crucial to establish this generalisation). So no need to be so dismissive right away. $\endgroup$
    – Vladimir Dotsenko
    Commented May 14, 2013 at 14:23

2 Answers 2

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Your expected generalization is true. Without loss of generality, you may assume that $G$ is a subgroup of $GL(V)$. A linear group whose elements have finite orders is locally finite, that's an old Theorem by Schur, and contained for instance in Wehrfritz's book Infinite Linear Groups. Now take a maximal linearly independent subset $S$ of $G$, so $\langle S\rangle$ and $G$ have the same invariant subspaces on $V$. By Schur, $\langle S\rangle$ is finite, so your generalization holds by Maschke.

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I. Schur explicitly proved that every periodic subgroup of ${\rm GL}(n,\mathbb{C})$ is completely reducible ( Corollary 36.3 in Curtis and Reiner's "Representation Theory of Finite Groups and Associative Algebras", Wiley and Sons, 1962), improving earlier results of Burnside,

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  • $\begingroup$ Even though your answer comes a bit late, it's always useful to provide a reference like this for a classical result rather than just sketching a proof. No need to reinvent the wheel here. $\endgroup$
    – Jim Humphreys
    Commented Dec 17, 2014 at 14:25
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    $\begingroup$ @JimHumphreys : I missed the question at the time, and just noticed it. I was motivated to give the reference to clarify what Burnside had done about subgroups of ${\rm GL}(n,\mathbb{C})$ (periodic such subgroups of bounded period are finite) and Schur had done ( finitely generated periodic such subgroups are finite and, in general, such periodic subgroups are completely reducible). $\endgroup$
    – Geoff Robinson
    Commented Dec 17, 2014 at 14:46

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