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While reading over Alperin's Local Representation Theory and reminding myself how a module is relatively H-projective iff H contains some vertex of the module, I realized I could not prove a basic lemma about relatively H-projective modules using my current favorite definition.

Definition: A G-module U is relatively H-projective if U is a direct summand of $(U_H)^G$.

Lemma: If HKG and U is a G-module that is relatively H-projective, then U is relatively K-projective.

Proof: Use Prop 9.1.2 to change definitions of relatively projective back to the good ole relative homological definition I used to like: H-split implies G-split. Clearly K-split implies H-split just by restricting the direct sum, but then relatively H-projectivity (Prop 9.1.2 style) gives G-split, but this suffices to show relative K-projectivity (Prop 9.1.2 style). $\square$

However, when I try to do this with $U|(U_H)^G$ I get confused and can't seem to use Mackey correctly. In Alperin's proof on page 67 he uses a similar but more convenient definition: that U is a direct summand of an induced H-module. Assuming my previous question is right, this is exactly the same as a direct summand of a relatively free module. However, I'm not sure I see how $U|(U_H)^G$ is equivalent to $U|S^G$ for some H-module S. Presumably this is Frobenius reciprocity, but again something is going wrong when I try to do it. At any rate, with the "S" definition it is just transitivity of induction, so I'd like to understand both.

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The key thing you need is that $S\uparrow^G $ is relatively $H$-projective in the following sense: for all surjections $M_1 \to M_2$ of $G$-modules, for all maps $\lambda: S\uparrow \to M_2$, if there's a map of $H$-modules $S\uparrow \to M_1$ making the diagram commute on restriction to $H$ then there's a map of $G$-modules making it commute. This follows from the universal property of induction.

Now apply this to the surjeciton $U\downarrow \uparrow \to U$ and $S\uparrow \to U$, where the first map is the natural surjection and the second is projection onto the $U$ factor. There is a $H$-map $S\uparrow \to U\downarrow \uparrow$ making this commute: project onto $U$, include this into $U\downarrow \uparrow$ by $u \mapsto 1\otimes u$. Thus there is a $G$-map $S\uparrow \to U\downarrow\uparrow$ making everything commute. This restricts to a $G$-map $U \to U\downarrow\uparrow$ splitting the natural surjection, so $U|U\downarrow\uparrow$.

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Thanks. Sorry I had misunderstood your answer earlier. This is quite clear to show the equivalence of U|U↓↑ and U|S↑. – Jack Schmidt Jan 27 '12 at 16:51

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