# How to prove $U \otimes Ind W = Ind(Res(U) \otimes W)$

How to prove $U \otimes Ind W = Ind(Res(U) \otimes W)$? where U is a representation of G and W is a rep of H, a subgroup of G. $Ind(W)$ is the induced rep and $Res(U)$ is the restrict rep.

Maybe I don't need a complete answer, so please just tell me, is the equal sign means being isomorphic? Are they the same meaning in rep theory, equal and isomorphic?

I got the answer, by both approaches: groups algebra and constructing isomorphic map. Thanks for all the very helpful comments and answer. I figured this is the wrong place to ask this question. This is the first time I ask here, I didn't know the rules, I apologize. I will come back after I know more math.

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Tell us your background and what books you have read on representation theory. –  KConrad Aug 22 '10 at 22:24
This is an exercise in Fulton and Harris, so without more information one might be tempted to think that this is "homework". –  José Figueroa-O'Farrill Aug 22 '10 at 22:31
Just me being ornery, but I strongly recommend you edit the question so that the title is a complete sentence, ending with a question mark, and the body doesn't begin mid-thought. As written, the question looks unprofessional. –  Theo Johnson-Freyd Aug 22 '10 at 23:00
The two are canonically isomorphic. (This is what the equality means.) Both $\otimes$ and Ind satisfy universal mapping properties; try combining them to figure out the universal property of each side of the purported isomorphism, and check that both satisfy the same universal mapping property, and hence are naturally isomorphic. –  Emerton Aug 23 '10 at 0:08
If the above comments aren't enough to help you, I recommend asking this on math.stackexchange.com as I think you'll get an answer more at the right level there. –  Andrew Stacey Aug 23 '10 at 10:28

Personally, I always find this one rather confusing. If you think in terms of modules over group rings, we want to show that $U \otimes (\mathbb{C}[G]\otimes_{\mathbb{C}[H]} W) \cong \mathbb{C}[G] \otimes_{\mathbb{C}[H]} (U\otimes W)$. The $G$-equivariant isomorphism is not given by sending $u\otimes (x\otimes w)$ to $x \otimes (u\otimes w)$. There's a lot wrong with this formula, but that's not the point.
Also, unless I'm confused, this really seems to depend on the structure of $\mathbb{C}[G]$ as both a ring and as a $\mathbb{C}[H]$-module. One needs to know that it's a free $\mathbb{C}[H]$-module, and that it has a decomposition as a $C[H]$-module into summands isomorphic to $C[H]$ that are permuted by the units of the ring $C[G]$. One could ask, for morphisms of rings $C\to R\to S$, when it's true that for an R-module M and an S-module N we have the formula $N \otimes_C (S\otimes_R M) \cong S \otimes_R (N \otimes_C M)$ (as S-modules). (Above I wrote $\otimes$ instead of $\otimes_{\mathbb{C}}$; now C is the ground ring). I don't know how to prove this without assuming S has the sort of structure mentioned above (free as an R-module, etc.).