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Let $G$ be a non-abelian group of order $2^5$ and center $Z(G)$ is non cyclic. Can we always find an element $x\not\in Z(G)$ of order $2$ if for any pair of elements $a$ and $b$ of $Z(G)$ of order $2$, the factor groups $G/\langle a\rangle$ and $G/\langle b\rangle$ are isomorphic?

Note: I do not find any contradiction of above statement so I wish to prove it.

user93432
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    Why don't you just do a computer search? There are just 35 groups of order $32$ with non-cyclic centres, so it should not take long! – Derek Holt Apr 28 '16 at 07:51
  • Thanks @DerekHolt for suggestion. I am trying. – user93432 Apr 28 '16 at 07:56
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    Of course, it might be more helpful and instructive to try and prove it by hand! It depends on the problem. If your hand proof is just a long case-by-case analysis, then a computer proof is likely to be more reliable, but an elegant hand proof is nicer! – Derek Holt Apr 28 '16 at 10:36

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This is not complete answer; but a partial information, which in addition to Holt's comment may simplify your job. (I will try to write complete proof as I get some directions on it)

Suppose $G$ is a group of order $2^5$ satisfying conditions in your question. We show that $Z(G)=C_2\times C_2$.

For this, by hypothesis, $|Z(G)|=2^2$ or $2^3$. If $|Z(G)|=2^3$, then $G/Z(G)=C_2\times C_2$. Hence there exists $x,y\in G$ be such that $G=\langle x,y,Z(G)\rangle$. Then $G'=\langle [x,y]\rangle$ and order of this commutator subgroup must be $2$ (fact: if $G/Z(G)$ is abelian then $G/Z(G)$ and $G'$ have same exponent.) Note that $G/Z(G)$ is abelian implies $G'\subseteq Z(G)$.

Since $Z(G)$ is non-cyclic, it should contain a subgroup of order $2$ other than $G'$; call it $N$. Then $G/N$ is non-abelian, whereas $G/G'$ is abelian. Thus, we are getting two subgroups of order $2$ in $Z(G)$ with non-isomorphic quotient.

Thus $|Z(G)|$ must be $4$ and it should be then $C_2\times C_2$ (it is non-cyclic).

p Groups
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  • Thanks @p Groups. Please provide me a reference for the fact that derived subgroup and inner automorphism group have same exponent if inner automorphism group is abelian. – user93432 Apr 29 '16 at 07:10
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    Suppose $G$ is a $p$-group such that $G'\subseteq Z(G)$. Then $[x,y^n]=[x,y]^n$ for all $x,y\in G$. This can be proved by the commutator identity $[a,cb]=[a,b][a,c]^b$ (see https://en.wikipedia.org/wiki/Commutator) and use induction. In argument(answer) above, note that since $G/Z(G)=C_2\times C_2$, so $G'\subseteq Z(G)$ hence exp($G'$)=exp($G/Z(G)$). – p Groups Apr 29 '16 at 16:20