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Let $G$ be a fnite group and $H$ a subgroup of $G$. Let $U$ be a $kG$-module and $V$ a $kH$-module. There is a natural isomorphism of $kG$-modules:

$U \otimes_k (Ind_H^G(V) ) \cong Ind_H^G(Res_H^G(U) \otimes_k V)$ sending $u \otimes (x\otimes v)$ to $x \otimes (x^{-1}u \otimes v)$ where $u \in U, x \in G$ and $v \in V$.

I cannot see how this is a $kG$-homomorphism. So when I do the following: $f(gu \otimes (x \otimes v)) = x \otimes (x^{-1}gu \otimes v)$, the $g$ and $x$ cannot move over the tensor, right? it just can't equal $gf(u \otimes (x \otimes v)) = gx \otimes (x^{-1}u \otimes v)$.

Any help would be appreciated!

scsnm
  • 1,283
  • When $M, N$ are $kG$, modules, then the action on $M\otimes_k N$ usually is diagonal, i.e. $g(m\otimes n)=(gm)\otimes(gn)$. Using this, one ahould have $gx$ in place of $x$ on the left hand side, and the $G$-equivariance follows. – Pavel Čoupek Sep 04 '20 at 04:39
  • clear! thx! so much! – scsnm Sep 04 '20 at 06:37

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