Let $G$ be a finite group and let $Z(G)$ denote its center. A simple result states that if $G/Z(G)$ a nontrivial cyclic group then $G$ is abelian. Of course if $G$ is abelian then $Z(G)=G$ and consequently $G/Z(G)$ is the trivial group. Hence we conclude that no non-trivial cyclic groups can result when we quotient a group by its center. My question is whether there are any other restrictions as to what groups can occur as the quotient of some group by its center.
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2"A simple result states that if $,G/Z(G),$ is cyclic non trivial then $,G,$ is abelian" ... – DonAntonio Apr 27 '13 at 23:56
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5But if $G/Z(G)$ is trivial, then $G=Z(G)$, hence is Abelian. – Berci Apr 28 '13 at 02:48
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This answer proves that $G/Z(G)$ cannot be $Q_8$. Also, this answer proves that the order of $G/Z(G)$ can be every powerful number (every prime factor has multiplicity $\ge 2$). Of course here one doesn't need to care about what the actual group given by the answer is, because there are so many $p$-groups of given order in general. – Jianing Song Aug 28 '23 at 17:45
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A group that can be written as $G/Z(G)$ for some group $G$ is called 'capable'. It appears to be easy to find necessary conditions for this, but difficult to find sufficient ones. The problem of finding such groups has been deemed 'interesting'. The question of which finitely generated abelian groups are capable was answered by Baer in 1938. Check out this, from our own @Arturo Magidin.
Alex Petzke
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@BabakS. I believe I did come across this idea originally from seeing it on Arturo's webpage, and it stuck with me. I too wish he would come back. – Alex Petzke Apr 28 '13 at 16:20