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Let $G=PQ$ where $P$ and $Q$ are $p$- and $q$-Sylow subgroups of $G$ respectively. In addition, suppose that $P\unlhd G$, $Q\ntrianglelefteq G$, $C_G(P)=Z(G)$ and $C_G(Q)\neq Z(G)$, where $Z(G)$ is the center of $G$.

I want to prove there exist two elements $x,y\in G-Z(G)$ such that $\left|C_G(x)\right| \nmid \left|C_G(y)\right|$ and $\left|C_G(y)\right| \nmid \left|C_G(x)\right|$.

By $C_G(Q)\neq Z(G)$, we know that there exists an element in $G$ which centralizes $Q$. By $C_G(P)=Z(G)$ we obtain that there is no non-central element that centralizes $P$. But if we can find an element that centralizes a big $p$-subgroup and small $q$-subgroup, we're done.

By GAP I have checked all groups of order less than $383$ with this hypothesis and couldn't find any counterexamples.

But I can't prove it!

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Adeleh
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  • If there is a few things that are useful but not perfect, I will greatful. – Adeleh Jul 31 '13 at 15:31
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    Previous questions in this series have been answered: http://math.stackexchange.com/questions/445042/c-gx-in-a-solvable-group http://math.stackexchange.com/questions/433576/a-question-about-solvable-group – Jack Schmidt Feb 09 '14 at 23:52
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    No counterexamples $G$ with $|G|\leq 1500$. – Jack Schmidt Feb 10 '14 at 11:25
  • I'm feeling this has something to do with group actions. We can use conjugacy classes and that'd give us a relation between the order of centralizers and the group. I'm going to give this a try! – Sum-Meister Feb 06 '17 at 14:35
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    An unsolved question so old and so well-received, I'm surprised it hasn't been cross-posted to MO. –  Aug 14 '18 at 03:15

1 Answers1

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You may be interested to note that if your conjecture is true then so is the apparently stronger conjecture:-

Let $G=PQ$ where $P$ and $Q$ are $p$- and $q$-Sylow subgroups of $G$ respectively, such that $P\unlhd G$ and $C_G(Q)$ is neither $Z(G)$ nor $G$. Then there are elements $x,y\in G$ such that neither of $|C_G(x)|$ and $|C_G(y)|$ is a factor of the other.

Proof

Suppose your conjecture is true and let $G=PQ$ be a group where $P$ and $Q$ are $p$- and $q$-Sylow subgroups of $G$ respectively, such that that $P\unlhd G$ and $C_G(Q)$ is neither $Z(G)$ nor $G$.

If $Q\unlhd G$, then $G$ is the direct product of $P$ and $Q$. If there were elements $x\in P-Z(P)$ and $y\in Q-Z(Q)$ then $|Q|$ would be a factor of $|C_G(x)|$ but not of $|C_G(y)|$ and $|P|$ would be a factor of $|C_G(y)|$ but not of $|C_G(x)|$. So we can suppose that either $P$ or $Q$ is abelian. If $P$ is abelian then $C_G(Q)$ is the direct product of $P$ and $Z(Q)$ but then we have the contradiction $C_G(Q)=Z(G)$. If $Q$ is abelian then we have the contradiction $C_G(Q)=G.$ We therefore conclude that $Q\ntrianglelefteq G$.

If $C_G(P)\ne Z(G)$, then let $x\in C_G(P)- Z(G)$ and $y\in C_G(Q)- Z(G)$. Then $|P|$ would be a factor of $|C_G(x)|$ but not of $|C_G(y)|$ and $|Q|$ would be a factor of $|C_G(y)|$ but not of $|C_G(x)|$. We therefore conclude that $C_G(P)=Z(G)$.

We now have all the conditions for the original conjecture and so there are two elements $x,y\in G$ such that neither of $|C_G(x)|$ and $|C_G(y)|$ is a factor of the other.