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Prove that there is no group $G$ such that the commutator subgroup of $G$ to be $S_{3}$.

maryam
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    Please note that most people here consider rude to write a question that looks like it consists only of a verbatim quote from an exercise sheet with no thought of the asker's own to go with it. What did you do to try to solve this problem? Where did you come upon it? – A.P. Apr 18 '13 at 19:08
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    See http://math.stackexchange.com/questions/361582/which-groups-are-derived-subgroups/361645#361645 – Derek Holt Apr 18 '13 at 20:35

1 Answers1

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Denote the commutator subgroup of a group $G$ by $G'$.

Since $S_3$ is complete, you could proceed as in this question. Alternatively, you could use the following idea:

If $G'/G''$ and $G''$ are cyclic, then $G'' = 1$.

Hints for proof: Show that $G''$ is in the center of $G'$. Conclude that $G'/Z(G')$ is cyclic, which implies that $G'$ is abelian.

Community
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Mikko Korhonen
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