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Suppose $G$ is a finite non-abelian group. Is it true that if $32 \nmid |G|$, then $G$ has at least one abelian centralizer?

Thanks in advance.

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    $\begingroup$ First of all, I don't have the faintest idea what the answer to your question is, but: what's this question's background? How did you get to it? What have you done/checked/tried?\ $\endgroup$ – DonAntonio Oct 18 '13 at 6:07
  • $\begingroup$ The smallest non-abelian group with no abelian centralizer is of order 32. Next is of order 64 and then 90. All these groups are divisible by 32. When I study the Centralizers of groups of order less than or equal to 100, I found that the answers to the above questions is yes. But I just can think where to start to prove, and to disprove I have to look on groups of order more than 100 and I'm still looking. $\endgroup$ – D. N. Oct 18 '13 at 7:45
  • $\begingroup$ Did you mean order $96$? Did you check the structure of the examples, as it seems probable that the "extra" ones would just be direct products of small groups with the example of order $32$. $\endgroup$ – Tobias Kildetoft Oct 18 '13 at 7:51
  • $\begingroup$ @TobiasKildetoft: Oh yes. My mistakes. Thank you. $\endgroup$ – D. N. Oct 18 '13 at 7:55
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    $\begingroup$ Try an extraspecial group of order 243. $\endgroup$ – Derek Holt Oct 18 '13 at 7:59
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As Derek Holt suggests in the comments, an extraspecial group of order $243 = 3^5$ is a counterexample. More generally, for any odd prime $p$, an extraspecial group of order $p^{2n+1}$ ($n \geq 2$) is a counterexample to your statement.

Proof: Let $|G| = p^{2n+1}$ be extraspecial, so $Z(G) = G'$ has order $p$. If $x \in Z(G)$, then $C_G(x) = G$ is nonabelian. If $x \not\in Z(G)$, the centralizer $C_G(x)$ has index $p$, since every conjugate of $x$ is contained in $xG'$. Suppose that $C_G(x)$ is abelian. Now for $y \not\in C_G(x)$, the intersection $C_G(x) \cap C_G(y)$ is central of index $p^2$, which implies $n = 1$ since $Z(G)$ has order $p$.

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