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The three subgroups of order $2$ in $S_{3}$ are conjugate and each is self-normalizing

Example

The subgroups

$⟨ (1 2) ⟩, ⟨ (1 3) ⟩, ⟨ (2 3) ⟩$

are the three subgroups of order $2$ in $S_{3}$ . They form one conjugacy orbit, and each equals its own normalizer.

Facts & Assumptions

Given: The symmetric group $S_{3}$ and $H = ⟨ (1 2) ⟩$ .

[L1]

The conjugates of $H$ are counted by $[S_{3} : N_{S_{3}} (H)]$ (The conjugates of $H$ are in bijection with $G / N_{G} (H)$ and, for finite $G$ , number $[G : N_{G} (H)]$ ).

[L2]

The normalizer consists of the elements preserving $H$ under conjugation (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

[L3]

The symmetric group consists of all permutations (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L4]

The symmetric group is a group under composition ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[L5]

The identity, the three transpositions, and the two $3$ -cycles exhaust the six elements of $S_{3}$ (The class equation of $S_{3}$ is $6 = 1 + 2 + 3$ ).

Verification

technique · direct

1.1

Each transposition generates a two-element subgroup, and these are distinct. Every element of order $2$ in $S_{3}$ is a transposition, so these are all the subgroups of order $2$ .

L3L4algebra

2.1

Conjugation relabels the two moved points, so the three subgroups in step 1.1 are conjugate.

step 1.1L2L3

3.1

By [L1], $[S_{3} : N_{S_{3}} (H)] = 3$ . Since $∣ S_{3} ∣ = 6$ by [L5], the normalizer has order $2$ ; it contains $H$ , so it equals $H$ . The same holds for each conjugate subgroup.

step 1.1step 2.1L1L2L5algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

P. Brosnan, Undergraduate Algebra Notes, 3.14: G-Sets, Example 3.111 (standard reference, not scraped)

The three subgroups of order 22 in S3S3​ are conjugate and each is self-normalizing