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.,

$$ S_3 ' = &lt;[a,b]:a,b, \in S_3&gt; .$$

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?
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 $S_3' = A_3$.
Instead of trying to find videos of other people working it out, why not try to work it out yourself?

After all, $S_3$ is a group of size 6. Here is a complete list of its elements: $(1),(12),(23),(13),(123),(132)$. Do you know how to "multiply" permutations? Do you know how to identify the "inverse" of a permutation? Then why don't you just try to plug in some of these elements for $a,b$ and figure out which element $[a,b]$ corresponds to? This might be slow for you at first, it should easily answer your question for whether $a,b$ need be distinct or if the order matters for $S_3$. More importantly, the exercise will deepen your basic understanding of groups.

EDIT: In general the question of "finding the commutator subgroup" is not well posed. In mathematics, the same object may be presented in many different ways. For example, in the case of $S_3$, the commutator subgroup is a cyclic subgroup of order 3. Thus, I could say the commutator subgroup is $C_3$, or I could say it's generated by $(123)$. For $S_3$ it's simple enough that either of these answers is pretty satisfactory. In general for $S_n$ it's still not bad - the commutator subgroup is a normal subgroup, so it's either the trivial group, $A_n$, or the whole group $S_n$. It's easy to rule out the first case by giving an example of a nontrivial commutator. You can rule out the last case by noting that a commutator, as a product of four permutations, must be an even permutation. Thus, the commutator subgroup must be $A_n$. Note that here, I'm using the fact that the only nontrivial proper normal subgroup of $S_n$ is $A_n$.

But what if I gave you a group $G$ defined to be the group generated by $x,y,z$ under the relations $x^3 = y^3z^2, y^5=1$, and $y^2 = x^2z^7$ what do you really mean by "finding the commutator subgroup?". In this case I suppose you could find generators, but that would be very hard to do without a computer (and possibly impossible).

Here's another example - Take $F_2$, the free group of two generators $x$ and $y$. What is its commutator subgroup? Well I guess you could just start trying to write down things of the form $aba^{-1}b^{-1}$, but of course there are infinitely many of them. How do you write them all down? In this case the subgroup is also infinitely generated, so if you only want a list of generators, how do you know that your list is complete? (of course this isn't That hard and there are papers giving an explicit list of generators, but most results in this case use topology). On the other hand, I can give you a very easy description of the commutator subgroup:

It is the subset of $F_2$ consisting of words whose exponents of $x$ sum to 0, and whose exponents of $y$ sum to 0. Why? Because the abelianization of $F_2$ is $\mathbb{Z}^2$ (this is easily checked using the fundamental theorem of finitely generated abelian groups). The abelianization map sends $x$ to $(1,0)$ and $y$ to $(0,1)$, and it's easily seen that the kernel consists precisely of words satisfying the above condition.

Anyway, in a sense you ARE missing the point. The whole point of being "rigorous/precise" in doing mathematics is so that we may take advantage of abstraction in order to deduce truths in extremely complicated (and often infinite) situations where explicit lists and formulas do not exist or do not help.

The key point is that you shouldn't shy away from quotient spaces or "abelianness". The language isn't there to confuse you - it's there to help you in cases where you can't just write down all the elements of the group and just "look at it". However, the interplay between the language and the explicit examples is also a good source of understanding. I would encourage you to work out lots of small examples, NOT for the purposes of circumventing the use of abstract theorems, but to help you understand the ideas underlying those theorems.