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Let $G$ be a group, $N, M$ normal subgroups with $N \cap M = {1}$ and $G = NM$. I know $N$ is a characteristic subgroup of $G$. How could I show that $M$ is characteristic as well?
Thank you.

P.S.: I also know that G is Abelian, but perhaps this fact isn't needed!?

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Well, if $\,G\,$ is abelian then any subgroup is normal, so why is that even mentioned? –  DonAntonio Dec 5 '12 at 11:49
    
Because the statement would be a lot stronger if that wasn't needed. –  Boris Dec 5 '12 at 12:00
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This isn't true. For example, consider $G=\mathbb{Z}\times\mathbb{Z_2}$. $0_{\mathbb{Z}}\times \mathbb{Z}_2$ is a characteristic subgroup of $G$, but $\mathbb{Z} \times 0_{\mathbb{Z}_2}$ is not.

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I'm sorry, but I don't see why $\mathbb{Z} \times 0_{\mathbb{Z}_2}$ is not characteristic. Could you help me? –  Boris Dec 5 '12 at 14:11
    
@Boris There is an automorphism of $G$ that doesn't preserve $\mathbb{Z}\times 0_{\mathbb{Z}_2}$. This automorphism sends $(a, b)$ to $(a, b + [a])$ for every $a \in \mathbb{Z}$ and every $b \in \mathbb{Z}_2$. Here $[a]$ denotes the class of number $a$ modulo $2$. –  Dan Shved Dec 5 '12 at 15:28
    
Thank you Dan. I understand it now. –  Boris Dec 9 '12 at 13:13
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