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Let $V$ be an $L$-module. I want to show that $V$ is a direct sum of irreducible $L$-submodules if each $L$-submodule of $V$ possesses a complement.

I want to show this via induction on the dimension of $V$. Do I start with $\dim V=1$ or $\dim V=2$ for my base case?

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Surely you need to assume "finite dimensional" somewhere... The proof must include the case $\dim V = 1$, though in that case the claim is trivial. Whether you need to do the case $\dim V = 2$ separately (as a special case) or not will depend on the precise argument in your inductive step; sometimes the inductive step requires the $n=2$ case to be already established, which is why it is proven separately. Sometimes it doesn't. –  Arturo Magidin Mar 1 '12 at 17:28
    
@Arturo, thanks for the edit! I will incorporate this style next time. –  Edison Mar 1 '12 at 17:28

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According to Joe's comment, a 1 dimensional module is a trivial sum of irreducible modules, to yes, start at the 1 dimensional case

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A $1$-dimensional module can be viewed as a trivial sum of irreducible modules. –  Joe Johnson 126 Mar 1 '12 at 17:30

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