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The three subgroups of order $2$ in $S_3$ are conjugate and each is self-normalizing

Example

The subgroups

$\langle(1\,2)\rangle,\qquad\langle(1\,3)\rangle,\qquad\langle(2\,3)\rangle$

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 $H=\langle(1\,2)\rangle$ .

[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 $N_G(H)=\{g\in G:gHg^{-1}=H\}$ of a subgroup).

[L3]

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

[L4]

The symmetric group is a group under composition ( $\operatorname{Sym}(X)$ 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

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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 222 in S3S_3S3​ are conjugate and each is self-normalizing