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CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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If [G:N][G:N] is finite then G/N=[G:N]|G/N|=[G:N]; for finite GG this equals G/N|G|/|N|

Statement

Let NGN\mathrel{\trianglelefteq}G. If [G:N][G:N] is finite, then the quotient group G/NG/N is finite and

G/N=[G:N].|G/N|=[G:N].

In particular, if GG is finite, then

G/N=GN.|G/N|=\frac{|G|}{|N|}.

Facts & Assumptions

Given: A group GG and a normal subgroup NGN\mathrel{\trianglelefteq}G.

[F1]

The index [G:N][G:N] is the finite cardinality of the left-coset set G/NG/N when that set is finite (The coset set G/HG/H and the index [G:H][G:H] of a subgroup).

[L1]

If GG is finite and NGN\le G, then G=[G:N]N|G|=[G:N]|N| (Lagrange's theorem: G=[G:H]H|G|=[G:H]|H| for every subgroup HH of a finite group GG).

Proof

technique · direct
1.1

If [G:N][G:N] is finite, then by [F1] the coset set underlying G/NG/N is finite with cardinality [G:N][G:N]; hence [F2] and [L2] give G/N=[G:N]|G/N|=[G:N].

F1F2L2
2.1

If GG is finite, then [L1] gives G=[G:N]N|G|=[G:N]|N|. Since NN contains the identity, N0|N|\ne0, and step 1.1 yields G/N=[G:N]=G/N|G/N|=[G:N]=|G|/|N|.

step 1.1L1algebra
3.1

The two asserted formulas follow.

step 1.1step 2.1

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Dependency tree · next 3 levels

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