TITLE: Question about Ext
QUESTION [8 upvotes]: I heard that $Ext(M,N)$ is naturally isomorphic to $Ext(M^*\otimes N,1)$ where 1 is the trivial representation and $M,N$ some representations of a group $G$.
Can anyone explain why?
Is there an explicit construction of a map from one to the other or does it just follow from some general considerations about derived functors?
Thanks.

REPLY [2 votes]: This is probably typographic, but the original question writes

Ext(M,N)  is naturally isomorphic to Ext(M*⊗N,1)

when what is true (see other answers) is that Ext(M,N) is isom.
to Ext(M $\otimes$ N*,1).
By the way, it is often useful to view Ext(M,N)
as Ext(1,M* $\otimes$ N) instead, since this later ext-group
coincides with the group cohomology H(G,M* $\otimes$ N);
here H*(G,-) = derived functor of "G-fixed-points".