S2 is not normal in S3

This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup (namely, Cyclic group:Z2 (?)) not satisfying a particular subgroup property in a particular group or type of group (namely, Symmetric group:S3 (?)).

Statement

Let $G$ be the symmetric group:S3, which is the group of all permutations on $\{1,2,3\}$. The group has the following six elements, given by their cycle decompositions:

$\!\{(),(1,2),(2,3),(1,3),(1,2,3),(1,3,2)\}$

Let $H$ be the subgroup $\{(),(1,2)\}$. In other words, $H$ is the symmetric group of degree two embedded inside $G$.

Then, $H$ is not a normal subgroup of $G$.

To explore more, see element structure of symmetric group:S3 and subgroup structure of symmetric group:S3.

Proof

Using conjugation

Consider the element $g=(2,3)\in G$ and its action by conjugation on $h=(1,2)\in H$. Since $g$ has order two, it equals its inverse, so $ghg^{-1}=(2,3)(1,2)(2,3)=(1,3)$. The element $(1,3)$ is not in $H$. Thus, $H$ is not normal in $G$.

Using cosets

Further information: S2 in S3#Cosets

The left cosets of $H$ in $G$ are:

$\!\{(),(1,2)\},\{(2,3),(1,3,2)\},\{(1,3),(1,2,3)\}$

The right cosets of $H$ in $G$ are:

$\!\{(),1,2)\},\{(2,3),(1,2,3)\},\{(1,3),(1,3,2)\}$

We see that the space of left cosets does not match the space of right cosets.

Using commutators

Consider the element $g=(2,3)\in G$ and $h=(1,2)\in H$. The commutator $[g,h]=ghg^{-1}h^{-1}$ is $(2,3)(1,2)(2,3)(1,2)=(1,2,3)\notin H$. Thus, $H$ is not normal in $G$.