# submodules of completely reducible modules

Let $L$ be a lie algebra, and $V$ be an $L$ module. Suppose that $V$ is completely reducible $L$ module (that is $V$ can be written as a direct sum of irreducible $L$ submodules of $V$). Is it necessarily true that if $U$ is a non-zero $L$ submodule of $V$, then $U$ must also be a completely reducible $L$ module? my answer is yes, but i am not really sure how to prove it.

Here is my attempt:

If $U$ is any $L$ submodule of $V$, then we can find an $L$ submodule of $V$ (say $U'$) such that $V=U\bigoplus U'$. Then $V/U'\simeq U$, so it suffices to show that $V/U'$ is completely reducible $L$ module. Let $\pi:V\rightarrow V/U'$ be a projection map. Write $V=\bigoplus V_{i}$, where each $V_{i}$ is an irreducible $L$ submodule of $V$. It suffices to show that each $\pi(V_i)$ is either zero or irreducible. This is the part that I am stuck. I am not sure if the assertion is true. In fact i am not sure if my approach/my assertion is also correct.

$\textbf{Edit}$: if my approach is correct, i would also like to know how to prove that each $\pi(V_i)$ is either zero or irreducible.

• This is true; but how do you get $U'$ with $V=U\oplus U'$? – Lord Shark the Unknown Mar 10 '18 at 7:14
• there is a result/theorem which says that $V$ is completely reducible if and only if each submodule of $V$ has a complementary submodule. – KnobbyWan Mar 10 '18 at 7:17
• there is also a result/theorem which says that every submodule of a completely reducible module is completely reducible. – Lord Shark the Unknown Mar 10 '18 at 7:19
• Why does it suffice to show that each $\pi(V_i)$ is either zero or irreducible? – JDZ Jul 10 '18 at 17:11

$\pi(V_i) \cong V_i / \ker(\pi|_{V_i})$ and $\ker(\pi|_{V_i})$ is either trivial or all of $V_i$ since $V_i$ is irreducible. The result follows.