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Let $N$ be normal subgroup of $G$ and $G=(N\times C_{3})\rtimes C_{2}$. Then prove $G=N\times (C_{3}\rtimes C_{2})$. Thank you

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What have tried Maryam? According to @Alexander's answer your claim in wrong? Are you sure about the conditions on $G$? Didn't you miss any assumptions? – S. Snape Dec 26 '12 at 9:54
May I suggest you this topic?… – Ivan Dec 26 '12 at 10:56

This is false. Let $N$ be any group for which $\text{Aut}(N)$ is even and let $C_2$ act trivially on $C_3$ and faithfully on $N$.

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I don't see it so clearly...perhaps an example would be a good idea. If I understood correctly, you mean that $\,C_3\rtimes C_2=C_3\times C_2\,$ , as the action is trivial, whereas $\,N\rtimes C_3\,$ is a non-trivial, and thus non-abelian, group...but then I think both groups are isomorphic. – DonAntonio Dec 26 '12 at 12:49
I meant that when $G=(N\times C_3)\rtimes C_2$, $C_2$ can act faithfully on $N$ and trivially on $C_3$, yielding $(N\rtimes C_2)\times C_3$. This is not in general isomorphic to $G=N\times (C_3\rtimes C_2)$, in which $C_2$ centralizes $N$. – Alexander Gruber Dec 26 '12 at 16:45
As a specific example we could take $N$ to be the cyclic group of order 5, and $G$ to be the direct product of the dihedral group of order 10 with a cyclic group of order 3. – Derek Holt Dec 26 '12 at 17:39

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