TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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

Statement

Let $H\le G$ have finite index $[G:H]=n$ , and put $N=\operatorname{Core}_G(H)$ . Then $N\mathrel{\trianglelefteq}G$ , the index $[G:N]$ divides $n!$ , and there are only finitely many subgroups $K$ with $H\le K\le G$ .

Facts & Assumptions

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

[L1]

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

[L2]

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

[L3]

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

[L4]

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

[L5]

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

[L6]

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

[L7]

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

[L8]

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

[L9]

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

[L10]

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

Proof

technique · direct

1.1

By [L1] and [L2], the coset action has kernel $N\mathrel{\trianglelefteq}G$ with $N\le H$ , and its image is a subgroup of $\operatorname{Sym}(G/H)$ .

L1L2L4

2.1

By [L3], $G/N\cong\operatorname{im}\rho$ . The set $G/H$ has $n$ elements by [L10], so [L5] gives $|\operatorname{Sym}(G/H)|=n!$ ; [L6] therefore gives $|G/N|=|\operatorname{im}\rho|\mid n!$ , that is, $[G:N]\mid n!$ by [L10].

step 1.1L3L4L5L6L10

3.1

Every subgroup $K$ containing $H$ also contains $N$ by [L2]. If $\pi(x)\in\pi[K]$ , then $\pi(x)=\pi(k)$ for some $k\in K$ , so $k^{-1}x\in N\le K$ and hence $x\in K$ ; thus $K=\pi^{-1}[\pi[K]]$ . Therefore $K\mapsto\pi[K]$ injects the set of such overgroups into the power set of the now known finite set $G/N$ , which is finite by [L8] and [L9].

step 2.1L2L7L8L9

4.1

Thus the core is normal and finite-index, its index divides $n!$ , and the collection of subgroups containing $H$ is finite.

step 1.1step 2.1step 3.1∎

Depends on

Used by

Every subgroup of index p in a finite p-group is normal Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 118 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

K. Conrad, Group Actions, Theorem 6.8 (standard reference, not scraped)

If [G:H]=n<∞[G:H]=n<\infty[G:H]=n<∞, then Core⁡G(H)⊴G\operatorname{Core}_G(H)\trianglelefteq GCoreG​(H)⊴G, [G:Core⁡G(H)]∣n![G:\operatorname{Core}_G(H)]\mid n![G:CoreG​(H)]∣n!, and only finitely many subgroups contain HHH