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

Let $$G=PQG=PQ$$ where $$PP$$ and $$QQ$$ are $$pp$$– and $$qq$$-Sylow subgroups of $$GG$$ respectively. In addition, suppose that $$P⊴P\unlhd G$$, $$Q\ntrianglelefteq GQ\ntrianglelefteq G$$, $$C_G(P)=Z(G)C_G(P)=Z(G)$$ and $$C_G(Q)\neq Z(G)C_G(Q)\neq Z(G)$$, where $$Z(G)Z(G)$$ is the center of $$GG$$.

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

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

By GAP I have checked all groups of order less than $$383383$$ with this hypothesis and couldn’t find any counterexamples.

But I can’t prove it!

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

Let $$G=PQG=PQ$$ where $$PP$$ and $$QQ$$ are $$pp$$– and $$qq$$-Sylow subgroups of $$GG$$ respectively, such that $$P\unlhd GP\unlhd G$$ and $$C_G(Q)C_G(Q)$$ is neither $$Z(G)Z(G)$$ nor $$GG$$. Then there are elements $$x,y\in Gx,y\in G$$ such that neither of $$|C_G(x)||C_G(x)|$$ and $$|C_G(y)||C_G(y)|$$ is a factor of the other.

Proof

Suppose your conjecture is true and let $$G=PQG=PQ$$ be a group where $$PP$$ and $$QQ$$ are $$pp$$– and $$qq$$-Sylow subgroups of $$GG$$ respectively, such that that $$P\unlhd GP\unlhd G$$ and $$C_G(Q)C_G(Q)$$ is neither $$Z(G)Z(G)$$ nor $$GG$$.

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

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

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