S2 is not normal in S3

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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.