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Let $G$ be a group. Let $Z(G)$ be the center of $G$, the set of elements that commute with every element of $G$.

Then, can we say that there is some elements in $Z(G/Z(G))$ which is not $Z(G)$?

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The sets $Z(G/Z(G))$ and $Z(G)$ are disjoint... they do not even share the identity element! – Mariano Suárez-Alvarez Apr 19 '12 at 16:28
@Mariano: "which is not", not "which is not in". $Z(G)$ is the identity element of $G/Z(G)$. – Chris Eagle Apr 19 '12 at 16:29
Oooooooooh. ${}$ – Mariano Suárez-Alvarez Apr 19 '12 at 16:38
Recall that any finite $p$-group has nontrivial center, so let $G$ be a nonabelian $p$-group... – Qiaochu Yuan Apr 19 '12 at 16:50

We can say it if we like, but it isn't true in general. Simple examples: if $G$ is abelian then $G/Z(G)$ is trivial and so has no non-identity elements. If $Z(G)$ is trivial, then $G/Z(G)$ is isomorphic to $G$ and again has no non-identity central elements.

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"in general"? How can I prove it? – lekrig Apr 19 '12 at 16:42
@lekrig: I don't understand what you are asking. The statement you ask about is false. I have proved that it is false by giving counterexamples. – Chris Eagle Apr 19 '12 at 17:08

Let's rephrase your question as follows:

Can we say that there is an element in $G$ whose image is in $Z(G/Z(G))$, but which is not in $Z(G)$?

Not always. In the two extreme cases, $G=Z(G)$ and $Z(G)=\{1\}$, the answer is "no". And there are examples of groups in which $Z(G)\neq \{1\}$, but $G/Z(G)$ has no center, so that the answer again is "no". For instance, take $G=S_3\times C_2$. Then $Z(G) = \{1\}\times C_2$, $G/Z(G)\cong S_3$, so $Z(G/Z(G))$ is trivial.

But there are some cases where we can guarantee it. Specifically, if $G$ is nilpotent but not abelian, then the answer is "yes". In particular, if $G$ is a nonabelian $p$-group, then the answer is "yes": there are always nontrivial elements in $Z(G/Z(G))$.

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The answer to this question depends on $G$. If $G$ is abelian, then there's no chance, since $G/Z(G)=\{Z(G)\}$.

On the other hand, consider the dihedral group given by symmetries of the square, whose center consists just of the identity and rotation by $\pi$. Any nontrivial coset will be in the center of $G/Z(G)\cong\mathbb{Z}/2\times\mathbb{Z}/2$, which is abelian, so in this case the answer is yes.

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I think you misunderstood my question. not "which is not in Z(G)", "which is not Z(G)" – lekrig Apr 19 '12 at 16:48
I did misread it, but I then edited my response. Is what I have now helpful? – Brett Frankel Apr 19 '12 at 16:50

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