Poincare's theorem

Statement

If a group (possibly infinite) has a subgroup of finite index, say $n$, then that subgroup contains a normal subgroup of finite index, where the index is at most $n!$. Specifically, we can take the normal core of the subgroup of finite index that we start with, as our normal subgroup of finite index.

Related facts

• Conjugate-intersection index theorem

Proof

In group action language

Given: A group $G$, a subgroup $H$ of index $n$

To prove: $H$ contains a subgroup $N$ that is normal in $G$, with index at most $n!$

Proof: Consider the action of $G$ by left multiplication on the coset space $G/H$. This gives a homomorphism from $G$ to the symmetric group $Sym(n)$. The kernel is precisely the intersection of the isotropies of all the points of $G/H$, or equivalently is the intersection of all conjugates of $H$. Call this $N$.

($N$ is also called the normal core of $H$).

Now, we have an injective map from $G/N$ to $Sym(n)$, so the index of $N$ is at most $n!$.