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Prove that M,N characteristic in G implies that MN is characteristic in G.

How do I prove this? Not sure where to start.

2 Answers 2

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$M,N\,\operatorname{char} G\Longrightarrow \phi(M)=M\,,\,\phi(N)=N\,\,\,\,\forall\,\phi\in\operatorname{Aut}(G)$

so

$\forall\,mn\in MN\,\,,\,\phi(mn)=\phi(m)\phi(n)\in MN\Longrightarrow \phi(MN)\subset MN$

and this is enough (why?)

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    Ahhh I get it now. Thanks!2012-11-12
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Here's a start: If $\phi$ is an automorphism of $G$, then for any $mn \in MN$ (where $m \in M$ and $n \in N$) we have $\phi(mn) = \phi(m)\phi(n)$. Try to show that $\phi(MN) = \phi(M)\phi(N)$ and use this to prove the result.