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Is it really true that there are no infinite irreducible representations of a finite group?

I was thinking a lot about this question but I haven't found any such representations.

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Let $A$ be a finite dimensional algebra (e.g. $A=kG$ the group algebra of a finite group which is the case you are asking about).

Let $L$ be an irreducible representation, then, there is a surjective map $A\to L$ from the regular representation (just map $1$ to a non-zero element). In particular, $L$ is finite dimensional.

The question about indecomposables is more difficult and was answered by Auslander in [Auslander: Representation theory of Artin algebras II. Comm. Alg. 1974] and [Auslander: Large modules over Artin algebras. In: Algebra, Topology and category Theory, Academic Press 1976]

Recall that a finite dimensional algebra is of finite representation type if there are only finitely many indecomposable modules up to isomorphism.

As Auslander has proven, if an algebra is of finite representation type, then any indecomposable module is finite dimensional and conversely, if a finite dimensional algebra is not of finite representation type, then there exists an infinite dimensional indecomposable module.

In the case of the group algebra (or more general a block of a group algebra). $kG$ (or a block $B$) is of finite representation type if and only if the characteristic $p$ of the ground field $k$ is zero or does not divide the order of the group (in which case the group algebra is semisimple and irreducible and indecomposable is the same) or the Sylow $p$-subgroups are cyclic (or in the case of a block $B$, the defect group of $B$ is cyclic).

For an explicit example, let $G=\mathbb{Z}/(2)\times \mathbb{Z}/(2)$ and $k$ be a field of characteristic $2$. Then $kG\cong k[x,y]/(x^2,y^2)$, an explicit isomorphism is given by sending the standard generators $(1,0)$ and $(0,1)$ of the Klein four group to $x+1$ and $y+1$. This is a special biserial algebra. For those, some infinite dimensional representations can actually be constructed, see e.g. [Krause: A note on infinite string modules]. For example take the vector space $\bigoplus_{i\in \mathbb{Z}} kv_i$ and define an action via $x$ sending $v_{2i}$ to $v_{2i-1}$ and $y$ sending $v_{2i}$ to $v_{2i+1}$ and all other actions being zero. This is indecomposable. The half-infinite versions of the module are also indecomposable.

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  • $\begingroup$ Thank you very much! Why is there such an surjective map $A\to L$ and where do you use that $L$ is irreducible? And can you name me a concrete group $G$ (not a group algebra) with an infinite dimensional indecomposable representation? $\endgroup$ – user404105 Oct 25 '17 at 17:01
  • $\begingroup$ Take any non-zero element $m\in L$. The morphism defined by $g\mapsto gm$ is surjective as its image is a non-zero submodule of $L$. For an explicit group you can take any non-cyclic $p$-group, E.g. A dihedral group. $\endgroup$ – Julian Kuelshammer Oct 25 '17 at 17:55
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    $\begingroup$ I answered it here. $\endgroup$ – Julian Kuelshammer Oct 26 '17 at 9:45
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    $\begingroup$ It might be a good idea to point out exactly where the isomorphism of algebras comes from (as it is not clear from the standard basis of the group algebra). $\endgroup$ – Tobias Kildetoft Oct 26 '17 at 12:04
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    $\begingroup$ Probably you can get an infinite dimensional indecomposable representation for any group by inducing one from its non-cyclic Sylow $p$-subgroup (I didn't think too hard about whether it will be indecomposable itself, but it should not decompose into finite-dimensional reps). $\endgroup$ – Tobias Kildetoft Oct 26 '17 at 19:04

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