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LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 GG be a finite group and HGH\le G. Then HH, every left and right coset of HH, and the coset set G/HG/H are finite. Moreover every coset has cardinality H|H|, and [G:H]=G/H[G:H]=|G/H| is a natural number.

Facts & Assumptions

Given: A finite group GG and a subgroup HGH\le G.

[L3]

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

[F1]

A bijection transports finiteness and finite cardinality (The cardinality A\lvert A\rvert of a finite set).

[F2]

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

Proof

technique · direct
1.1

Since HGH\subseteq G and GG is finite, HH is finite by [L1].

givenL1
1.2

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

F2L1L2
2.1

Every coset is equinumerous with HH, so every coset is finite and has cardinality H|H|.

step 1.1L3F1
3.1

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

step 1.1step 2.1step 1.2F2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 71 results over 26 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