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(Related to this question.)

Suppose that $H$ is a finite subgroup, and $g$ is an element of a group. How large can the intersection $gH\cap Hg^{-1}$ be given that $gH\ne Hg^{-1}$? Is it possible to have, say, $\frac{1}{2}|H|<|gH\cap Hg^{-1}|<|H|$?

The abelian case is easy: if the group is commutative, then $gH$ and $Hg^{-1}$ coincide if $g^2\in H$, and are disjoint otherwise.

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2 Answers 2

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We cannot have $|H|/2 < |aH \cap Hb| < |H|$ for any $a,b \in G$.

Since $g \in aH \Rightarrow gH=aH$, this would imply $|H|/2 < |gH \cap Hg| < |H|$ for some $g \in G$. But $|gH \cap Hg| = |H \cap g^{-1}Hg|$, and $H \cap g^{-1}Hg$ is a subgroup of $H$, so it cannot satisfy that inequality.

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  • $\begingroup$ Corrected (should be $gH=aH$) $\endgroup$
    – Derek Holt
    Feb 12 at 13:40
  • $\begingroup$ Sorry, I still do not understand how $Hb$ becomes $Hg$. Posting my own solution meantime... $\endgroup$
    – W-t-P
    Feb 12 at 13:53
  • $\begingroup$ If $g \in aH \cap Hb$ then $aH=gH$ and $Hb = Hg$. $\endgroup$
    – Derek Holt
    Feb 12 at 14:05
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Suppose that $|gH\cap Hg^{-1}|>\frac12\,|H|$. Then, by the box principle, there is an element $h\in H$ such that $gh=h^{-1}g^{-1}$. We have then $gHg=ghHg=h^{-1}g^{-1}Hg$. As a result, $|gH\cap Hg^{-1}|=|gHg\cap H|=|h^{-1}g^{-1}Hg\cap H|=|g^{-1}Hg\cap H|$, and since $g^{-1}Hg\cap H$ is an intersection of two subgroups of size $|H|$, the size of the intersection is either $|H|$, or $\frac12\,|H|$ at most.

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