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

Let G=PQ where P and Q are p– and q-Sylow subgroups of G respectively. In addition, suppose that P, 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!

Answer

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.

Attribution
Source : Link , Question Author : Adeleh , Answer Author : Community

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