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?
2
$\begingroup$
abstract-algebra
group-theory
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0Well, if $\,G\,$ is abelian then any subgroup is normal, so why is that even mentioned? – 2012-12-05
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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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0I'm sorry, but I don't see why $\mathbb{Z} \times 0_{\mathbb{Z}_2}$ is not characteristic. Could you help me? – 2012-12-05
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0@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 $2$. – 2012-12-05
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0Thank you Dan. I understand it now. – 2012-12-09