Finding the commutator subgroup of $S_3$ and the commutator subgroup of that commutator subgroup.

I'm finding this idea very frustrating. I need to find the commutator subgroup of $S_3$ (call it $S_3'$) and the commutator subgroup of that commutator subgroup (call it $S_3 ''$).

So, I have that, for $S_3$, the commutator of $a,b \in S_3$ is given by

$$ [a,b] :=aba^{-1}b^{-1}. $$

And the commutator subgroup is the subgroup generated by all such $[a,b]$, i.e.,

$$ S_3 ' = <[a,b]:a,b, \in S_3> .$$

Ok, I have read about everything I can on this and watched every single YouTube video on the subject I can find, and have yet to see it just worked out explicitly in a way that makes sense to me. For example: these elements $a,b$. Do they need to be distinct from one another?! Is this an ordered pair, or not?! I know that I only need to check the normal subgroups, of which there are only three, but that's still a lot of things to check if they do not need to be distinct and need to be ordered. I just feel like there are a lot of basic questions here that I'm not seeing addressed anywhere in the literature, which makes me feel like I'm "missing the point," in a way.

EDIT: for $S_3 ' = \{e, (123), (132) \}$ here are all of my computations in an attempt to determine $S_3 ''$:

$$ [e, (123)] = e \\ [e, (132)] = e \\ [(123, (132)] = e \\ [(132), (123)] = e \\ [(123), e] = e \\ [(132), e] = e \\ $$

Ok. So, certainly the subgroup generated by $e$ is simply $e$, which means we get to conclude that $S_ 3 '$ is abelian? Is that it?

1 Answer

A priori $a$ and $b$ need not be distinct, but if $a = b$ then $[a, b] = [a, a] = e$, and so such choices contribute nothing to the subgroup. Applying the definition shows that $[a, b] = [b, a]^{-1}$, so commutator in general depends on the ordering.

Hint For this particular group, note that regardless of the parity of $a, b$, the commutator $[a, b]$ has even parity. So, $S_3'$ is generated by some subset of $A_3$ and hence $S_3' \leq A_3$. At this point, it becomes useful to compute one or more commutators manually.

Additional hint Computing the commutator of the transpositions $(12)$ and $(13)$ gives that $S_3'$ contains $$[(12), (13)] = (12)(13)(12)^{-1}(13)^{-1} = (12)(13)(12)(13) = (123) .$$ This $3$-cycle has order $3$ and so generates $A_3$, so $S_3' \geq A_3$ and thus $S_3' = A_3$.