= Commutator subgroup of the symmetric group on three letters
{title2=$S_3'=A_3$}
For the <symmetric group> $S_3$, take the <transpositions> $x=(12)$ and $y=(23)$. Products of <permutations> apply the rightmost factor first. Then
$$
xy=(123),\qquad yx=(132),\qquad [x,y]=(12)(23)(12)(23)=(132).
$$
The last <permutation> sends $1\mapsto3$, $3\mapsto2$, and $2\mapsto1$: two swaps fail to commute by an entire three-cycle.
\Image[/s3-commutator-and-abelianization.png]
{title=<Group commutators> and the information surviving <abelianization> in the <symmetric group> $S_3$}
{description=The upper rows group the six <permutations> by <sign of a permutation>. Each triangle labels the original vertices in their new positions. The lower row applies $y,x,y^{-1},x^{-1}$ successively, giving $[x,y]=(132)$ under the rightmost-first convention. The <commutator subgroup> consists of the three <even permutations>; the <center of a group> here contains only $e$.}
Every <group commutator> has <sign of a permutation> $+1$, so $S_3'\subseteq A_3$, the <alternating group> of <even permutations>. Conversely, the displayed <group commutator> generates $A_3=\{e,(123),(132)\}$. Therefore $S_3'=A_3$. The two <cosets> are the <even permutations> and the <odd permutations>, and their multiplication is just multiplication of signs:
$$
S_3^{\mathrm{ab}}=S_3/A_3\cong\{+1,-1\}\cong C_2.
$$
This is the <first isomorphism theorem> applied to the <sign of a permutation>. Once <commutativity> is forced, all three <transpositions> become indistinguishable, and both three-cycles become the <identity element>. The surviving information is exactly whether the <permutation> is even or odd. Compare the very different <center of a group> $Z(S_3)=\{e\}$.
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