CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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For $K \leq H \leq G$ with $G$ finite, $[G : K] = [G : H] [H : K]$

Statement

If $K \leq H \leq G$ and $G$ is finite, then all three indices are finite and

$[G : K] = [G : H] [H : K] .$

Facts & Assumptions

Given: A finite group $G$ and subgroups $K \leq H \leq G$ .

[L1]

Lagrange's theorem gives $∣ B ∣ = [B : A] ∣ A ∣$ whenever $A \leq B$ and $B$ is finite (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , The coset set $G / H$ and the index $[G : H]$ of a subgroup, The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[F1]

Every subgroup contains the identity; hence its underlying set is nonempty. Also, if $K \leq H \leq G$ , then $K \leq G$ : one has $K \subseteq H \subseteq G$ , and the identity, product, and inverse conditions for $K$ are the same inherited operations in $H$ and $G$ (Subgroup).

[F2]

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

[L2]

Natural multiplication is associative, and $x z = y z$ with $z \neq 0$ implies $x = y$ (Multiplication is associative, Cancellation for multiplication by a nonzero factor).

Proof

technique · direct

1.1

Since $H \leq G$ , its underlying set is a subset of the finite set $G$ , so $H$ is finite by [F2]. Also $K \leq G$ by the subgroup-transitivity derivation in [F1]. Applying [L1] to $K \leq H$ , $H \leq G$ and this $K \leq G$ gives $∣ H ∣ = [H : K] ∣ K ∣$ , $∣ G ∣ = [G : H] ∣ H ∣$ , and $∣ G ∣ = [G : K] ∣ K ∣$ .

givenF1F2L1

2.1

Substituting the first equality into the second and comparing with the third gives $[G : K] ∣ K ∣ = ([G : H] [H : K]) ∣ K ∣$ .

step 1.1L2

3.1

Since $K$ contains the identity, $∣ K ∣ \neq 0$ . Cancellation in $N$ therefore yields $[G : K] = [G : H] [H : K]$ .

step 2.1F1L2∎

Depends on

Used by

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Thomas W. Judson, Abstract Algebra: Theory and Applications, Lagrange's Theorem (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §6.2: Lagrange's Theorem (standard reference, not scraped)

For K≤H≤GK≤H≤G with GG finite, [G:K]=[G:H][H:K][G:K]=[G:H][H:K]