Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02
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The three-cycle subgroup of Sym⁡({1,2,3}) is normal and its quotient has two elements

Example

In G=Sym⁡({1,2,3}), let N=⟨(123)⟩. Then

N={id⁡,(123),(132)}

is normal, and G/N is the two-element group {N,(12)N}.

Facts & Assumptions

Given: Permutations are composed from right to left.

[F1]

The subgroup generated by an element consists of its integral powers (The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups).

[F2]

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

[L2]

A subgroup of index two is normal (Every subgroup of index two is normal).

[L3]

A normal subgroup gives a quotient group under coset multiplication (For N⊴G, the cosets form a group with identity N and inverse (gN)−1=g−1N).

Verification

technique · direct
1.1

Since (123)2=(132) and (123)3=id⁡, [F1] gives N={id⁡,(123),(132)}.

L1F1algebra
2.1

The remaining three permutations are the transpositions, and (12)N={(12),(23),(13)}. Thus N and (12)N are the two left cosets of N in G.

step 1.1algebra
3.1

Consequently [G:N]=2 by [F2], so N⊴G by [L2].

step 2.1F2L2
4.1

The quotient group therefore exists by [L3], and its underlying coset set is precisely {N,(12)N}. Its order is two by [L4].

step 2.1step 3.1L3L4∎

Depends on

Used by

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Sources