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
Let $N$ be a normal subgroup of group $G$ and $G=(N\times C_{3})\rtimes C_{2}$. Then prove $G=N\times (C_{3}\rtimes C_{2})$.
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group-theory
finite-groups
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0What 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? – 2012-12-26
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0May I suggest you this topic?http://math.stackexchange.com/questions/264096/semi-direct-groups-isomorphisms – 2012-12-26
1 Answers
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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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0I 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. – 2012-12-26
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0I 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 $N$. – 2012-12-26
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0As 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. – 2012-12-26