Conjugate-intersection index theorem

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
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Statement

Symbolic statement

Let $H$ be a subgroup of $G$ of finite index $r$ in $G$. Let $H_{1},H_{2},\ldots ,H_{s}$ be distinct conjugates of $H$ in $G$, so that $s\leq r$. Let $K=\bigcap _{i=1}^{s}H_{i}$. Then:

• $K$ is of finite index in $G$
• The index of $K$ in $G$ is bounded from above by $r(r-1)\ldots (r-s+1)$ viz. $[r]_{s}$ or $P(r,s)$.

Related facts

• Poincare's theorem bounds the index of the normal core of a subgroup, in terms of the index of the subgroup

Proof

We have a natural action of $G$ on the coset space $G/H$ by left multiplication. Now, consider the set of $s$-tuples over $G/H$ with all entries distinct. $G$ acts on this set by the coordinate-wise action on $G/H$.

Now, for each conjugate $H_{i}$, find a $g_{i}\in G$ such that $H_{i}=g_{i}Hg_{i}^{-1}$.

Claim: The group $K=\bigcap _{i}H_{i}$ is the isotropy subgroup for the element $(g_{1}H_{1},g_{2}H_{1},g_{3}H_{1},\ldots g_{s}H_{1})$

Proof: Note that the isotropy subgroup for this point under the coordinate-wise action must stabilize every coordinate, and conversely, if a group element fixes every coordinate, it must belong to the isotropy subgroup. Hence, the isotropy subgroup is the intersection of the isotropy subgroups for every coordinate. But the isotropy subgroup for $g_{i}H$ is precisely $g_{i}Hg_{i}^{-1}=H_{i}$. This gives the result.