# Let $H\le G$ be finite and $g\in G$. How large can $gH\cap Hg^{-1}$ be if $gH\ne Hg^{-1}$? Is $\frac{1}{2}|H|<|gH\cap Hg^{-1}|<|H|$ possible?

(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.

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.
• Corrected (should be $gH=aH$) Feb 12 at 13:40
• Sorry, I still do not understand how $Hb$ becomes $Hg$. Posting my own solution meantime... Feb 12 at 13:53
• If $g \in aH \cap Hb$ then $aH=gH$ and $Hb = Hg$. Feb 12 at 14:05
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.