# Surjective map results in subrepresentation

I need to prove that a surjective homomorphism of finite $\mathbb{F}_p[\Delta]$-modules $$A \twoheadrightarrow B$$ results in $B$ being a subrepresenation of $A$ of the group $\Delta$ of order prime to $p$. I, however, don't know if this really leads to a subrepresentation, since I've always thought, that a subrepresentation should look like $B \subset A$ in module language. Thanks for your help, Tom

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Exactly how is the original exercise worded? And, is $\Delta$ any group? – Berci Apr 17 '13 at 16:43
a subrepresentation should look like $B\subseteq A$ in module language I agree with that. In any case, $B$ would be a quotient representation of $A$... – rschwieb Apr 17 '13 at 17:01
A surjection means $B$ is a quotient of $A$. Now $\Delta$ is prime to $p$, so representations are semisimple, so subrep and quotient rep is the same thing. – Aaron Apr 17 '13 at 19:47
Thank you for your hint! Are submodules and quotient modules the same in the semi-simple case because every short exact sequence splits? – BIS HD Apr 17 '13 at 21:05
The $k[G]$ modules are semisimple, so all submodules are direct summands. – tharris Apr 17 '13 at 22:29

## 1 Answer

As tharris and Aaron pointed out, for a field $k$ (e.g. $\mathbb{F}_p$) whose characteristic does not divide the group order $|\Delta|$, Maschke's theorem asserts that any $k[\Delta]$-module is semisimple, which is per definition equivalent to the fact that every submodule is a direct summand. Hence any quotient module must be a submodule.

Thanks to you, guys!

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