Question regarding Jordan Holder support and isotypic components.

Question

So for an $R$ module $M$ let $M_\pi$ be the sum of the simple submodules of $M$ with $\pi$ as their isomorphism class with $\pi \in Irr(R)$ - the set of equivalence classes of simple $R$ modules. I want to show that $(M_\pi)_{\pi \in Irr(R)}$ is linearly independent. To do so, I want to show that for a simple module $E,$ if $E \subset M_{\pi_1} + \ldots _+ M_{\pi_n}$ then the isomorphism class of $E$ belongs to $\{\pi_1, \ldots, \pi_n\}.$ To do so, first we can see that $E \subset E_1 + \ldots + E_m$ where the $E_i$ have isomorphism classes belonging to $\{\pi_1, \ldots, \pi_n\}.$ My professor says that the Jordan Holder support for $E_1 + \ldots + E_m$ is a subset of $\{\pi_1, \ldots, \pi_n\}.$ I can see this as $0 \subset E_1 \subset E_1 + E_2 \subset \ldots \subset E_1 + \ldots E_m$ is a composition series and at each step the isomorphism class of the quotient belongs in $\{\pi_1, \ldots, \pi_n\}.$ Then he says that the isomorphism class of $E$ belongs in the Jordan Holder support. Why is this? Is it because we can construct a composition series of $E_1 + \ldots + E_m$ beginning with $0 \subset E \subset \ldots?$ Any help is appreciated thanks.

Answer 1

Once you have a composition series $0 = A_0 < A_1 < A_2 < ... < A_n = M$, simply form the series $0 < E=A_0+E < A_1+E < A_2+E < ... < A_n+E=M$. Show that $A_{j+1}/A_j$ maps surjectively onto $(A_{j+1}+E)/(A_j+E)$, so each factor in the new series is either a composition factor, $E$, or $0$. But then omitting the $0$s (in fact there will be only one 0, which is easy to see), you have a series in which each factor module is simple, so it must be a composition series, and $E$ is one of these modules, so it must be isomorphic to a composition factor from the original series, by Jordan-Holder.