# A question about Sylow subgroups and $C_G(x)$

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!

• If there is a few things that are useful but not perfect, I will greatful. – Adeleh Jul 31 '13 at 15:31
• Previous questions in this series have been answered: math.stackexchange.com/questions/445042/… math.stackexchange.com/questions/433576/… – Jack Schmidt Feb 9 '14 at 23:52
• 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 6 '17 at 14:35
• An unsolved question so old and so well-received, I'm surprised it hasn't been cross-posted to MO. – Robert Wolfe Aug 14 '18 at 3:15

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.