TITLE: Compute the derived subgroup of $S_3$
QUESTION [0 upvotes]: I am trying to understand why $(S_3)' \cong A_3$
The resonning goes like this:
$S_3/A_3\cong C_2$
since $C_2$ commutes, $(S_3)' ≤ A_3$ (That I know from a theorem in the course).
I now need an explanation for the reasonning that follows
$A_3 \cong C_3$
Since $S_3$ does not commute ,$ (S_3)' \neq \{Id\} \implies (S_3)'  \cong A_3$
So more specifically here is where I have issues:

Why is $A_3 \cong C_3$ and why does it matter?
I understand that $(S_3)$ is commutative iff $(S_3)' = \{Id\}$
Now I understand the conclusion only if $A_3$ is simples in $S_3$ but I don't know if that's the case. I researched it but couldn't find an answer.

REPLY [3 votes]: It matters because $C_3\cong A_3$ implies that $A_3$ is very short on subgroups and $S_3'$ must be one of them.
Uhm... No? To me, making sense of that statement is rather challenging. One (but not the sole) of the reasons for this is the fact that you seem to confuse the verb "to commute" (such as in "$g\in G$ and $h\in G$ commute with each other" = "$gh=hg$") with "being commutative" (such as in "the group $G$ is commutative" = "all the elements of $G$ commute with each other").
After edit: Indeed, in general a group $G$ is commutative if and only if $G'=\{e\}$. In fact, if $G'\ne \{e\}$, there must be a commutator $ghg^{-1}h^{-1}\ne e$, id est $gh=ghg^{-1}h^{-1}hg\ne hg$. Viceversa, if $gh\ne hg$, then $ghg^{-1}h^{-1}\ne e$.
$A_3$ is simple. In general, $A_n$ is simple if and only if $n\ne 4$. As far as I know, simplicity is an intrinsic property (a group is simple if it has no normal subgroups), not a relative one: thus I wouldn't know how to interprete being "simple in $S_3$".
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