The inspiration for this came from these videos. Particularly part 4 in this series.

The S 3 symmetry group on three symbols.

Specifically the elements represent permutations of 3 objects such that e (identity element) = (1,2,3), τ 1 = (1,3,2), τ 2 = (2,1,3), τ 3 = (3,2,1), σ 1 = (3,1,2), σ 2 = (2,3,1). The multiplication operator is defined to be composition of permutations. That is a*b is defined as to mean permute the 3 objects according to b, then permute the resulting arrangement by a. When all the permutations are calculated in this way then the result must be one of the permutations listed above (as this is an exhaustive list). This gives the group multiplication table as:

Observations: The group is non-abelian. I then investigated normal subgroups of the form gHg-1 = H. (In the diagrams below the horizontal legend refers to elements in H while the vertical legend refers to elements in G.)

In this table there are 3 normal subgroups. Firstly the identity is in its own subgroup as g*e*g-1 = e always. This is not particularly surprising as I think this applies to all groups.

There is however one more normal subgroup here.

So both the (σ)s and (τ)s form a normal subgroup by themselves. However, one mystery I am trying to solve at the moment is the claim made in the video that all normal subgroups are the kernel of a group homomorhpism. The is understandable for the identity element (e) but I’m not sure why this is the case for the (σ)s and (τ)s.

Update: 3-June-2016

After watching the videos on quotient groups:

and

Now it finally makes a bit more sense. The equivalence classes which are also cosets are {e, σ 1 , σ 2 } and {τ 1 , τ 2 , τ 3 }. The first is a normal subgroup, the second coset is not. (For instance τ 1 * τ 2 = σ 2 and therefore the τ’s are not closed under *).