CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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

Statement

If $K\le H\le G$ 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 $K\le H\le G$ .

[L1]

Lagrange's theorem gives $|B|=[B:A]|A|$ whenever $A\le B$ 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 $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity).

[F1]

Every subgroup contains the identity; hence its underlying set is nonempty. Also, if $K\le H\le G$ , then $K\le G$ : 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 $\lvert B\rvert \le \lvert A\rvert$ , and equality holds if and only if $B = A$ ).

[L2]

Natural multiplication is associative, and $xz=yz$ with $z\ne0$ implies $x=y$ (Multiplication is associative, Cancellation for multiplication by a nonzero factor).

Proof

technique · direct

1.1

Since $H\le G$ , its underlying set is a subset of the finite set $G$ , so $H$ is finite by [F2]. Also $K\le G$ by the subgroup-transitivity derivation in [F1]. Applying [L1] to $K\le H$ , $H\le G$ and this $K\le G$ 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, $|K|\ne0$ . Cancellation in $\mathbb N$ therefore yields $[G:K]=[G:H][H:K]$ .

step 2.1F1L2∎

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: 84 results over 21 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

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\le H\le GK≤H≤G with GGG finite, [G:K]=[G:H][H:K][G:K]=[G:H][H:K][G:K]=[G:H][H:K]