S2 is not normal in 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.

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
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 $$G$$.