# Lemma $5.1$ of Elements of the Representation Theory of Associative Algebras

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?

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