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I'm reading “Elements of the representation theory of associative algebras” by Assem, Simson and Skowroński. I'm not convinced by the proof of lemma 5.1.a.

The statement is as follows:

Let $A$ be a finite-dimensional $K$-algebra and let $G$ be finite group and $H\leq G$ a subgroup. If the group algebra $AG$ is representation-finite, then $AH$ is representation-finite.

Proof: Denote by $M_1, \dots ,M_t$ all the indecomposable $AG$-modules (up to isomorphism). Considering each $M_i$ as a $AH$-module, we get that $M_i\cong \bigoplus_{j}N_{ij}$ where the $N_{ij}$'s are indecomposable $AH$-modules. (This follows from the unique decomposition theorem and the fact that we are considering only finitely generated modules). Now let $N$ be a indecomposable $AH$-module. Consider the map $p:AG\rightarrow AH:\sum_{g\in G}\alpha_g g\mapsto \sum_{h\in H}\alpha_hh$. Clearly this map is an epimorphism, a retraction and an $AH$-$AH$-bimodule map.

It follows that $N\otimes_{AH}AG\xrightarrow{1_N\otimes p}N\otimes_{AH}AH\cong N$ is a retraction as well, and thus $N$ is a direct summand of $N\otimes_{AH}AG$. The book then claims that $N\otimes_{AH}AG$ is isomorphic to "the direct sum of the modules $M_i$." The conclusion then is obvious.

However, I do not see why $N\otimes_{AH}AG$ would be isomorphic to the direct sum of the $M_i$'s. Also, since $N\otimes_{AH}AG$ is an $AH$-module, this isomorphism should be an $AH$-module isomorphism. And I would be very surprised to hear that $N\otimes_{AH}AG$ is such a sum independent of the indecomposable module $N$. What am I missing or is this nonsense?

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When they say that $N\otimes_{AH}AG$ is "the direct sum of the modules $M_i$", the use of the word "the" is confusing, but they just mean that as an $AG$-module, $N\otimes_{AH}AG$ is a direct sum of indecomposable modules, each of which is isomorphic to some $M_i$, and therefore as an $AH$-module it is a direct sum of indecomposable modules, each of which is isomorphic to some $N_{ij}$. Since $N$ is a direct summand (as an $AH$-module) of $N\otimes_{AH}AG$, it must be isomorphic to one of the (finitely many) modules $N_{ij}$.

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  • $\begingroup$ Yes that makes sense. I was also a bit confused by the structure on $N\otimes_{AH}AG$, but I deconfused myself. Thank you. $\endgroup$ – Mathematician 42 Oct 11 '16 at 12:12

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