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Let $k$ be a field and $G$ a group. Suppose we have $kG$-modules $A$ and $B$, and we want to consider $A\otimes_k B$ as a $kG$-module via $g(a\otimes b)=ga\otimes gb$. The module axioms are easy enough to verify, but how does one show this action is even well-defined?

Thoughts: Consider the map $\phi_g:A\times B\to A\otimes_k B$ given by $(a,b)\mapsto ga\otimes gb$, where $g$ is just a fixed group element. I want to show this is $kG$-balanced (so that we have a well-defined group homomorphism on $A\otimes_k B$). It's obviously additive in both coordinates, but it's not clear to me why $\phi_g(a,xb)=\phi_g(ax,b)$ for $x\in kG$. This seems to involve passing $g$ by $x$, but that is very suspicious

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The easiest way to proceed, in my mind, is to note that giving a $k$-linear $G$-action on a $k$-module $M$ is the same thing as giving a group homomorphism $G \to \textrm{Aut}_k (M)$. So each element $g$ of $G$ corresponds to an automorphism $\theta_g$ of $M$ and an automorphism $\psi_g$ of $N$. Since $\otimes_k$ is functorial, we get an automorphism $\theta_g \otimes_k \psi_g$ of $M \otimes_k N$, and it's not too hard to see that this defines a homomorphism $G \to \textrm{Aut}_k(M \otimes N)$. – Zhen Lin Sep 18 '12 at 5:03
up vote 2 down vote accepted

The tensor product here is the tensor product over $k$, not the tensor product over $k[G]$. You only need to show that $\phi_g$ is $k$-balanced, not $k[G]$-balanced (which won't be the case in general if $G$ is not abelian).

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Let $x=kg'$, then $\phi_{g}(a,xb)=ga\otimes gkg'b=kga\otimes gg'b$. While $\phi_{g}(ax,b)=gakg'\otimes gb=kgag'\otimes gb$. So you want to show $ga\otimes gg'b=gag'\otimes gb$. Unless you make your basic ring to be $k[G]$ you cannot have such a relationship. Qianchu's remark seems to be helpful in suggesting using $x\in k$ altogether.

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