Calculating the commutator (derived) subgroup of $S_3$

Question

How can I calculate it easily?

I showed that the commutator group of $S_3$ is generated by $(123)$ in $S_3$ using the fact that $S_3$ is isomorphic to $D_6$ and relation in $D_6$ but that was tedious

Are there any elegant methods?

Thanks.

Answer 1

The commutator subgroup can be characterised as being the smallest normal subgroup such that the quotient is abelian; by this I mean that if $N$ is normal in $G$ with $G/N$ abelian then $N$ contains the commutator subgroup.

The only normal subgroups of $S_3$ are $\{e\}$ and the $C_3$ subgroup generated by $(123)$ (being normal is a direct consequence of having index $2$).

The respective quotients have sizes $1$, $6$ and $2$ and since $S_3$ is nonabelian and a group of order $2$ must be, we can say that the commutator subgroup must be $C_3$.

Answer 2

If $x$ and $y$ are in $S_3$, then their commutator, $x^{-1}y^{-1}xy$, is an even permutation. So the commutator subgroup is a subgroup of $A_3$, which is just the identity and the 3-cycles.

Now since there are elements of $S_3$ that don't commute, the commutator subgroup isn't just the 1-element group, so it must be all of $A_3$.