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Andy Putman
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Your memory is correct, at least if you replace $\mathbb{Z}$ with a field $k$ of characteristic $0$. This is a theorem of Gaschütz. See

W. Gaschütz, Über modulare Darstellungen endlicher Gruppen, die von freien Gruppen induziert werden, Math. Z. 60 (1954), 274–286.

One way to prove it is to think of your free group as the fundamental groups of a graph $X$ with one vertex and $n$ loops. Then $R$ is the fundamental group of a regular cover $Y$ of $X$ with deck group $G$. Since the deck group acts freely on $Y$ its cellular chain complex is a chain complex of $G$-modules that can be identified with

$$0 \rightarrow k[G]^n \rightarrow k[G] \rightarrow 0.$$

When you take homology you know you get $k$ in degree $0$, and the theorem now follows from semisimplicity.

Your memory is correct. This is a theorem of Gaschütz. See

W. Gaschütz, Über modulare Darstellungen endlicher Gruppen, die von freien Gruppen induziert werden, Math. Z. 60 (1954), 274–286.

Your memory is correct, at least if you replace $\mathbb{Z}$ with a field $k$ of characteristic $0$. This is a theorem of Gaschütz. See

W. Gaschütz, Über modulare Darstellungen endlicher Gruppen, die von freien Gruppen induziert werden, Math. Z. 60 (1954), 274–286.

One way to prove it is to think of your free group as the fundamental groups of a graph $X$ with one vertex and $n$ loops. Then $R$ is the fundamental group of a regular cover $Y$ of $X$ with deck group $G$. Since the deck group acts freely on $Y$ its cellular chain complex is a chain complex of $G$-modules that can be identified with

$$0 \rightarrow k[G]^n \rightarrow k[G] \rightarrow 0.$$

When you take homology you know you get $k$ in degree $0$, and the theorem now follows from semisimplicity.

Source Link
Andy Putman
  • 44.8k
  • 14
  • 186
  • 272

Your memory is correct. This is a theorem of Gaschütz. See

W. Gaschütz, Über modulare Darstellungen endlicher Gruppen, die von freien Gruppen induziert werden, Math. Z. 60 (1954), 274–286.