# Poincare's 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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This article gives the statement, and possibly proof, of a subgroup property (i.e., subgroup of finite index) satisfying a subgroup metaproperty (i.e., normal core-closed subgroup property)
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## Statement

### 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).

## 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 $n!$. 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 $N \le H \le G$.