# The cardinality of the intersection of a subgroup and its conjugate

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