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!?
Two normal subgroups with trivial intersection, one is characteristic, what about the other?
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abstract-algebra
group-theory
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0Because the statement would be a lot stronger if that wasn't needed. – 2012-12-05
1 Answers
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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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0Thank you Dan. I understand it now. – 2012-12-09