Mathematics Asked by user792898 on December 21, 2021

Let $G$ be a finite group and $A$ be a subgroup of $G$. Suppose $A^x$ and $A^y$ are two conjugates of $A$ such that $A$, $A^x$ and $A^y$ are not equal. Do we always have $|Acap A^x|=|Acap A^y|$?

I tried to find a counterexample in $S_4$ but failed. Does this hold for any finite group and its subgroups? Any help is appreciated! Thanks!

A counterexample can be found for example in $G=S_3 times S_3$: put $A={((1),(1)),((1),(12)),((23),(1)),((23),(12))}$. Then $A$ is a Sylow $2$-subgroup of $G$. And so are $B={((1),(1)),((1),(13)),((23),(1)),((23),(13))}$ and $C={((1),(1)),((13),(13)),((13),(1)),((1),(13))}$. Of course, by Sylow Theory all $2$-Sylow subgroups are conjugated. But here $A cap B={((1),(1)),((23),(1))}$, while $Acap C={((1),(1))}$. So, $|A cap B|=2$ and $|A cap C|=1$.

Answered by Nicky Hekster on December 21, 2021

View $G$ as a transitive group acting on the set of cosets $Omega=[G:G_alpha]$, where we write $A=G_alpha$. Then begin{equation*} G_alphacap G_alpha^x=G_{(alpha,alpha^x)} end{equation*} is the point-wise stabiliser of ${alpha,alpha^x}$. Consider the action of $G_alpha$ on $Omegasetminus{G_alpha}$. The stabiliser of $alpha^x$ in $G_alpha$ is exactly $G_{(alpha,alpha^x)}$, the order of which is begin{equation*} G_{(alpha,alpha^x)}=frac{|G_alpha|}{|(alpha^x)^{G_alpha}|}. end{equation*}

If the length of orbits $|(alpha^x)^{G_alpha}|$ and $|(alpha^y)^{G_alpha}|$ containing $alpha^x$ and $alpha^y$ respectively are different then $|G_{(alpha,alpha^x)}|ne|G_{(alpha,alpha^y)}|$. Usually there are several subdegrees in a transitive action, and so suitable $x$ and $y$ are easily obtained. That means, your equation can only hold in very special cases.

Answered by Hongyi Huang on December 21, 2021

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