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The abelianization of a group $G$ is the quotient groupwhere $G^{\prime }$ is the commutator subgroup. Its elements are cosets, and its operation is $(g{G}^{\prime })(h{G}^{\prime })=gh{G}^{\prime }$. The result is an abelian group: the two products $gh$ and $hg$ differ by a group commutator, which disappears in the quotient group.

For the symmetric group $S_3$, the sign of a permutation realizes $S_3^{\mathrm{ab}}\cong C_2$, as shown in commutator subgroup of the symmetric group on three letters. For the dihedral group of a square, abelianization of the square symmetry group gives $C_2\times C_2$: two independent yes-or-no pieces of information survive. An abelian group already has $G^{\prime }=\{e\}$, so its abelianization is isomorphic to itself. At the other extreme, a perfect group has trivial abelianization.

Calling this the largest abelian quotient means that every group homomorphism into any abelian group factors uniquely through $q$. It does not require $G$ to be finite, or compare the sizes of unrelated abelian groups. The universal property of abelianization makes this precise. It lets us solve all such mapping problems in one simpler group: any invariant that respects multiplication and takes values in an abelian group sees only $G^{\mathrm{ab}}$.

The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.