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Let $M_R$ be a right $R$-module. A homogeneous component $H$ of $M_R$ is defined to be the sum $\sum_{i\in I}B_i$ where $\lbrace B_i \rbrace_{i\in I}$ is a family of mutually isomorphic simple submodules $B_i \subseteq M$. Is it true that if the homogeneous component $H$ is non-simple then $H$ is decomposable as $H=N\oplus K$ or $H=N \oplus K \oplus L$ where $N\cong K$ and $L$ is simple?

Thanks in advance.

Hussein Eid
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1 Answers1

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Yes.

$H$ is semisimple by definition, so you can write it as $\oplus_{i\in I}H_i$ for some index set $I$ and all the $H_i$'s are mutually isomorphic and simple.

Then either

  1. $H$ has finite composition length and it is

    1. even, in which case $N$ and $K$ are a sum of half the composition length copies of the simple module or
    2. odd, in which case you split off one copy $L$ of the simple module and divide the rest into two equal groups to get $N$ and $K$; or

  2. $H$ does not have finite composition length, in which case you can just divide the index set into two equipotent pieces and use them to find $K$ and $N$.

rschwieb
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  • This is what I have thought of. But I have two questions. Question 1: Where is the use of the hypothesis that $H$ is non-simple ?!. Question 2: What is the guarantee that the sum $N + K$ is direct (In other world, how to guarantee that $N \cap K = 0$ ?!). – Hussein Eid Nov 09 '21 at 00:03
  • @HusseinEid 1) I don’t see any use for it. It first the second case when $N=K={0}$. 2) the homogeneous component is semisimple by definition, so you can always express it as a direct sum of simples. – rschwieb Nov 09 '21 at 00:44