Does $Ind_{H}^{G}\mathbb{Q}\cong\mathbb{Q}[G/H]$?

Question

I know that $Ind_{H}^{G}\mathbb{Z}\cong\mathbb{Z}[G/H]$, but I am unsure whether $Ind_{H}^{G}\mathbb{Q}\cong\mathbb{Q}[G/H]$ would hold? I don't understand why it would but at the same time can't prove it.

Answer 1

First, $\mathrm{Ind}_H^G\mathbb{Q}=\mathbb{Q}[G]\otimes_{\mathbb{Q}[H]}\mathbb{Q}$ is spanned by the set $\{g\otimes 1\mid g\in G\}$ and, moreover, we have $$g\otimes 1 = g'\otimes 1\iff g^{-1}g'\in H.$$ (Note: $g^{-1}g'=h\in H\iff g'=gh\iff g'\otimes 1=gh\otimes 1=g\otimes(h.1)=g\otimes 1$)

Define a map $$\phi:\mathrm{Ind}_H^G\mathbb{Q}\to\mathbb{Q}[G/H]$$ by $\phi(g\otimes 1)=gH$. By the above paragraph, this is obviously a linear isomorphism. It is just left to observe that it is $G$-equivariant, which is essentially obvious by the definitions of the actions.