My question is about a problem that its assumptions are like Maschke's theorem in some ways.

Let $\mathbb{F}$ be a field that $char \mathbb{F}$ doesn't divide $|G:H|<\infty$ and $M$ be an $\mathbb{F}G$-module that is completely reducible $\mathbb{F}H$-module. We want to show that $M$ is completely reducible $\mathbb{F}G$-module

I know that the group $G$ in this problem must be infinite and with this assumption, it is far from Maschke's theorem, but with the asumption of $|G:H|<\infty$, I think that the proof can be like that of Maschke's theorem but I don't know how. I will be so appreciate if someone give me some hints or prove this statement here.


Suppose $i: N\rightarrowtail M$ is an ${\mathbb F}G$-submodule of $M$. Then, since $M$ is completely reducible over ${\mathbb F}H$, $i$ admits an $H$-linear retraction $\rho: M\twoheadrightarrow N$. Can you now adapt the proof of Maschke's theorem to average $\rho$ so as to make it even $G$-linear?

It might to help to recall that for the proof of Maschke's Theorem the crucial point is the existence of the element $\omega := \frac{1}{|G|}\sum_{g\in G} g\otimes g^{-1}$ in ${\mathbb F}G\otimes_{\mathbb F} {\mathbb F}G$ satisfying $g\omega=\omega g$ for all $g\in G$, and $\mu(\omega)=1$, where $\mu: {\mathbb F}G\otimes_{\mathbb F}{\mathbb F}G\to {\mathbb F}G$ is the multiplication; that is to say, ${\mathbb F}G$ is separable over ${\mathbb F}$. Try to mimic this to construct an element of ${\mathbb F}G\otimes_{{\mathbb F}H}{\mathbb F}G$ showing the separability of ${\mathbb F}G$ over ${\mathbb F}H$ in case $[G:H]<\infty$ and $[G:H]\neq 0$ in ${\mathbb F}$, and use this element to adapt the linearization of $\rho$ above.

More explicitly In the classical proof of Maschke's theorem, given an embedding $i: N\rightarrowtail M$ of ${\mathbb F}G$-modules, you first pick an ${\mathbb F}$-linear retraction $\rho: M\twoheadrightarrow N$ and afterwards consider the averaging $\widetilde{\rho}: M\to N$ defined by $\widetilde{\rho}(m) := \frac{1}{|G|}\sum\limits_{g\in G} g^{-1}.\rho(g.m)$. Then firstly $\widetilde{\rho}\circ i = \text{id}$, and $g_0^{-1}\widetilde{\rho}(g_0.m)=\widetilde{\rho}(m)$ follows from the fact that $g\mapsto gg_0$ is a permutation on $G$. Now, in your more general context, assume that $\rho$ is already $H$-linear and use a suitable sum over representatives of $H$-cosets in $G$ instead of summing over all elements of $G$. The permutation argument proving the $G$-linearity of $\widetilde{\rho}$ can then be generalized.

  • $\begingroup$ Dear Hanno, thank you very much for your answer. Is there any way to solve this problem without using the tensor product of modules? $\endgroup$ – user187409 Nov 3 '14 at 20:59
  • $\begingroup$ @Alice: Sure, I added some $\otimes$-free details. $\endgroup$ – Hanno Nov 3 '14 at 21:17
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    $\begingroup$ Notice that we dont need $M$ to be completely reducible as an $\mathbb{F} H$-module for this proof. We just need to find a complement that is invariant under $H$ for every submodule of $M$ as a $\mathbb{F}G$-module. This is clearly weaker. $\endgroup$ – kevkev1695 Dec 6 '19 at 16:56

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