Poincare's theorem

Verbal 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 $$n!$$. In fact, we can choose the normal subgroup such that its index is a multiple of $$n$$ and a divisor of $$n!$$.

(The subgroup that we choose here is the normal core of the original subgroup).

Symbolic statement
Suppose a group $$G$$, a subgroup $$H$$ of index $$n$$. Then, $$H$$ contains a subgroup $$N$$ that is normal in $$G$$, with index in $$G$$ at most $$n!$$. In fact, we can choose $$N$$ such that:

$$n|[G:N]|n!$$.

(The subgroup that we choose here is the normal core of the original subgroup).

Other facts about normal subgroups and index using the idea of the action on the coset space

 * Conjugate-intersection index theorem: This gives a bound on the intersection of finitely many conjugate subgroups
 * Subgroup of index two is normal
 * Subgroup of least prime index is normal

Analogous facts for characteristic subgroups

 * Replacement theorem by characteristic subgroup satisfying multilinear commutator identities

Facts used

 * 1) uses::Group acts on left coset space of subgroup by left multiplication
 * 2) uses::First isomorphism theorem
 * 3) uses::Lagrange's theorem
 * 4) uses::Index is multiplicative

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 $$n!$$. Further, we can choose $$N$$ such that $$n|[G:N]|n!$$.

Proof:


 * 1) Consider the action of $$G$$ by left multiplication on the left coset space $$G/H$$ (fact (1)). This gives a homomorphism from $$\varphi:G \to \operatorname{Sym}(n)$$.
 * 2) Let $$N$$ be the kernel of $$\varphi$$. Then $$N$$ is normal and $$N \le H$$: The kernel is precisely the intersection of the isotropies of all the points of $$G/H$$; equivalently, it is the intersection of all conjugates of $$H$$. In particular, $$N \le H$$. $$N$$ is also the normal core of $$H$$.
 * 3) The index of $$N$$ is at most $$n!$$. In fact, it divides $$n!$$: By the first isomorphism theorem (fact (2)), $$G/N$$ is isomorphic to the image $$\varphi(G)$$, which is a subgroup of $$\operatorname{Sym}(n)$$. Thus, the index of $$N$$ is at most $$n!$$. In fact, fact (3) (Lagrange's theorem) yields that the order of $$\varphi(G)$$ divides $$n!$$. Hence, the index $$[G:N]$$ also divides $$n!$$.
 * 4) The index of $$N$$ is a multiple of $$n$$: This follows from fact (4), applied to the groups $$N \le H \le G$$.

Textbook references

 * , Page 48, Exercise 20