Jordan-Holder theorem for modules? Let $A$ be a finite dimensional algebra over some field $k$.
I think from the Jordan-Holder Theorem, one might be able to claim that every simple $A$-module occurs in the series (by this I mean it is isomorphic to the quotient of two successive submodules in the composition series).
I would be thankful if anyone could help me with the following questions. 


*

*Is my thought true (about the occurrence of simple modules)?

*Is it possible to have two (possibly) different but isomorphic quotients in the composition series? In other words is it possible for a simple module to be isomorphic to more than one quotient of the successive submodules in the composition series.
I want to conclude from the above questions, whether there is a relation between the length of a composition series and the number of equivalences classes of simple $A$- modules?     
**By modules I mean left $A$-modules and by a series I mean the composition series of $A$.
 A: For 1
Yes, it's true. The trick is to remember that the simple modules of $A$ are the same as the simple modules of $A/J(A)$, where $J(A)$ is the Jacobson radical of $A$.
Since $A$ is a finite dimensional algebra, it is a right and left Artinian and Noetherian ring. As such, it has a composition as a left module over itself (and as a right module over itself too.) If we can show that every simple factor already appears in a composition series for $A/J(A)$, then we just link it up with a composition series for $J(A)$ and get a composition series for $A$ which contains copies of all isotypes of simple modules in its factors.
Now $A/J(A)$ is a semisimple ring, and all isotypes appear as factors in a composition series for $A/J(A)$. Can you see why this is? 
A basic approach to prove this would be to remember that all isotypes of simple left $A/J(A)$ modules appear as minimal left ideals. That makes it clear that they appear in a decomposition of $A/J(A)$ into a direct sum of simple modules. By chopping off one summand at a time, you can produce a composition series displaying all the isotypes in its composition factors.
For 2
Sure, isotypes can occur multiple times, and you may have already noticed this if you have already carried out the last paragraph. Take a semisimple ring $R$ and decompose it into simple left $R$ modules: $R=S\oplus S'\oplus T$ where $S\cong S'$ are isomorphic simple left ideals and $T$ is another simple left ideal nonisomoprhic to $S$ and $S'$. Then this is a composition series:
$$
\{0\}\subseteq S\subseteq S\oplus S'\subseteq S\oplus S'\oplus T=R
$$
The first two factors are isomorphic.
For the final question
Yes: your instinct is right. From this it follows that the composition length of $A$ is bounded from below by the number of distinct isotypes of simple $A$ modules. If you say a composition series $\{0\}=S_0\subseteq\ldots\subseteq S_n=A$ has length $n$, then $n$ is greater or equal to the number of distinct isoclasses.
A: Q1:  Every simple $A$-module is of the form $A/m$ for some maximal ideal $m$ of $A$(proof is easy).Now we can write(as $A$ is noetherian and artinian) a composition series $A\supset m \supset  \ldots \supset 0$  of  $A$. So $A/m $ is occurring in at least one composition series as a factor .Then Jordan-Holder asserts that $A/m$ occurs in any composition series.
