Number of submodules of a module with non-isomorphic composition factors Let $A$ be a finite dimensional algebra over $\mathbb{C}$ and let $M$ be a finite-dimensional $A$-module.
Suppose further that $M$ is semisimple with composition factors $S_1, \dots, S_k$, where $S_i \not\cong S_j$ for all $i \neq j$. I'm attempting to show that $M$ has only finitely many submodules.
By semisimplicity we can write $M = \bigoplus_{i=1}^k M_i$ and from the Jordan-Holder theorem applied to the composition series $0 \subset M_1 \subset M_1 \oplus M_2 \subset \ldots \subset M_1 \oplus M_2 \oplus \ldots \oplus M_k$ we can arrange that $M_i \cong S_i$ after reordering.
My suspicion is that any submodule is of the form $\bigoplus_{i \in I} M_i$ for some $I \subseteq \{1, \ldots, k\}$ so that there are $2^k$ submodules. I've tried a number of different ways to prove this but haven't quite been able to get the result I'm looking for.
Let $N$ be a submodule of $M$. My first thought was to consider a composition series for $M$ containing $N$. I think this gets us our result only up to isomorphism, which isn't quite strong enough.
My next thought was to consider the projection maps $\pi_i \colon N \to M_i$. If $\ker\pi_t = N$ then $N \subset \bigoplus_{i \neq t}M_i$ and the result follows by induction on $k$.
Now suppose that $\ker\pi_i \neq N$ for all $1 \leq i \leq k$. Because each $M_i$ is simple we then have $N / \ker \pi_i \cong M_i$ for each $i$. My hope was to show that if $N$ is too small then we must, by some counting argument, have that two of the kernels are the same, therefore contradicting the non-isomorphism of the $M_i$. I'd then conclude that $N = M$ and be done.
As one avenue, by altering our induction to be on $k + \dim N$ we can write each kernel as a direct sum of the $M_i$. My hope was that each kernel is then "determined" in some sense by the fact that it misses its corresponding simple module. I tried using the maximality of the kernels in $N$ but that didn't get me anywhere.
That's about as far as I've managed to get. Am I along the right lines with any of this or am I exploring a dead end? This problem came up in the context of an introductory representation theory course so maybe the machinery of representations could be helpful, but I can't see how.
 A: A bit of a late reply to this question, however I'm answering this for my benefit, more than anything else.
It's fairly intuitive to say that a direct sum of any combination of these $S_{i}$ is a submodule of M (we know we have $2^k$ of these because the set of combinations is just a power set of k elements, which always has that many elements.) What you need is that any submodule of M sits in a composition series of $M$, so can be represented as some direct sum of composition factors.
So, let $A$ be a $K$-algebra, let $V$ be a finite dimensional $A$-module, and let $L{\subset}V$ be a submodule.
Consider $V/L$ and $L$. If both are simple, then we have that
${0} \subset L \subset V$
is a composition series of $V$. If they're both zero we have a trivial composition series for $V$. If they are not simple we find some module $W$ such that
${0} \subset L \subset W \subset V$.
If $L$, $W/L$ and $V/W$ are simple then we have done it. Otherwise we keep taking refinements in this manner (if L is non-simple we can take the composition series for that and stick it in, in the manner of the proof of the Jordan Holder theorem) until we have our simple composition factors. We know this procedure terminates because V is finite dimensional.
So then applying this we have that all submodules of M must lie in a composition series, and as such must be of this form.
A: I came across this question when trying to solve the very same problem. As mentioned by another user, the problem with the answer above is that it proves that there are finitely many up to isomorphism. You can actually refine the argument and prove that there are finitely many.
We first prove that it has finitely many simple submodules. Let $N \subseteq M$ be simple. Then, the projections $\pi_i: N \to M$ are all zero except for one $i$, since they are pairwise non-isomorphic. But then, $N = M_i$.
Now suppose $N$ is not simple, and pick a simple module $N' \subseteq N$ (this is possible due to the finite dimension). Then, $N' \cong M_i$ for some $i$. It is not too difficult to prove that then
$$ N \cong M_i \oplus \left( N \cap \bigoplus_{j \neq i} M_j \right) $$
and the argument now follows by induction.
