If $[G:H]=n<\mathrm{\infty }$, then ${\mathrm{Core}}_G(H)⊴G$, $[G:{\mathrm{Core}}_G(H)]\mid n!$, and only finitely many subgroups contain $H$

Given: A group $G$, a subgroup $H\le G$ of finite index $n$, and $N:={\mathrm{Core}}_G(H)$.

The action on $G\mathrm{/}H$ gives a homomorphism $\rho :G\to \mathrm{Sym}(G\mathrm{/}H)$ whose kernel is $N$ (Left multiplication on $G\mathrm{/}H$ is transitive, has stabiliser $H$ at $H$, and has kernel ${\mathrm{Core}}_G(H)$).

The core $N$ is normal in $G$, lies in $H$, and contains every normal subgroup of $G$ lying in $H$ (${\mathrm{Core}}_G(H)$ is the largest normal subgroup of $G$ contained in $H$).

First isomorphism gives $G\mathrm{/}\ker \rho \cong \mathrm{im}\rho$ (First isomorphism theorem for groups: $G\mathrm{/}\ker f\cong \mathrm{im}f$).

The symmetric group of a set is its group of bijections (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition, $\mathrm{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

A set with $n$ elements has exactly $n!$ bijections to itself (A finite set $A$ with $\mid A\mid =n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality, The factorial $n!$ and the falling factorial $n^{\underset{‾}{k}}$, defined by recursion in $\mathbb{N}$).

The order of a subgroup of a finite group divides the order of the group (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).

The canonical projection $\pi :G\to G\mathrm{/}N$ is a surjective homomorphism (The canonical projection $\pi :G\to G\mathrm{/}N$, $\pi (g)=gN$, is a surjective group homomorphism, The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$).

The power set of a finite set is finite ($\mid \mathcal{P}(A)\mid =2^{\mid A\mid }$ for finite $A$).

Every subset of a finite set is finite (A subset of a finite set is finite, with $\mid B\mid \le \mid A\mid$, and equality holds if and only if $B=A$).

For a finite-index subgroup, the index is the cardinality of its coset set (The coset set $G\mathrm{/}H$ and the index $[G:H]$ of a subgroup, The cardinality $\mid A\mid$ of a finite set).