LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-01

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In a finite group, the subgroup, every coset and the set of cosets are finite

Statement

Let $G$ be a finite group and $H \leq G$ . Then $H$ , every left and right coset of $H$ , and the coset set $G / H$ are finite. Moreover every coset has cardinality $∣ H ∣$ , and $[G : H] = ∣ G / H ∣$ is a natural number.

Facts & Assumptions

Given: A finite group $G$ and a subgroup $H \leq G$ .

[L1]

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]

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

[L3]

Every left or right coset of $H$ is equinumerous with $H$ (Every left or right coset of $H$ is equinumerous with $H$ ).

[F1]

A bijection transports finiteness and finite cardinality (The cardinality $∣ A ∣$ of a finite set).

[F2]

The coset set is $G / H = {g H : g \in G}$ , and its finite cardinality is the index (The coset set $G / H$ and the index $[G : H]$ of a subgroup, The left cosets of a subgroup partition the group).

Proof

technique · direct

1.1givenL1

Since $H \subseteq G$ and $G$ is finite, $H$ is finite by [L1].

1.2F2L1L2

Every coset is a subset of $G$ , so $G / H \subseteq P (G)$ . The power set is finite by [L2], hence $G / H$ is finite by [L1].

2.1step 1.1L3F1

Every coset is equinumerous with $H$ , so every coset is finite and has cardinality $∣ H ∣$ .

3.1step 1.1step 2.1step 1.2F2∎

Therefore $[G : H] = ∣ G / H ∣ \in N$ , and the finiteness and cardinality assertions are steps 1.1, 1.2 and 2.1.

In a finite group, the subgroup, every coset and the set of cosets are finite

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources