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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 $G^{\mathrm{ab}}$.

For odd $n\ge 3$, the dihedral group $D_{2n}$ of order $2n$ has rotation generator $r$ and reflection generator $s$, with $srs=r^{-1}$. The group commutator $[r,s]=r^{-2}$ generates the rotation subgroup, while the quotient by that subgroup is $C_2$. Hence the commutator subgroup is exactly $⟨r⟩$ and the abelianization is $C_2$. The only abelian quotients are $C_2$ and the trivial group.

Let $D_8$ denote the dihedral group of group order eight acting on a square. Here the subscript specifies group order; the same group is called $D_4$ in the convention that counts vertices of a polygon, as in the group action example. Let $r$ be a counterclockwise quarter-turn rotation and $s$ a reflection in a diagonal. ThenProducts act rightmost first. The group commutator of the generators of a group isIt is a half-turn rotation, so forcing $r$ and $s$ to commute must identify a half-turn with $e$.

More concretely, write each element uniquely as $r^i{s}^j$, with $0\le i<4$ and $j\in \{0,1\}$. The functionis a surjective group homomorphism. Indeed, $srs=r^{-1}$ giveswhere the exponents are reduced modulo four and two. Modulo two, $-k$ and $k$ agree, so $\varphi$ adds the two coordinates separately. Its group kernel is $N=\{e,r^2\}$. Since the target is an abelian group, $D_8^{\prime }\subseteq N$; since $[r,s]=r^2$, also $N\subseteq D_8^{\prime }$. ThusThe four cosets are $\{e,r^2\}$, $\{r,r^3\}$, $\{s,r^{2}s\}$, and $\{rs,r^{3}s\}$. Each coset forgets the difference of a half-turn. The quotient group still distinguishes rotations from reflections, and even from odd numbers of quarter-turns; the two surviving choices are independent. This is the Klein four-group, rather than the cyclic group $C_4$, because each nonidentity element squares to the identity element. For the general dihedral group calculation, see Conrad, Dihedral groups I, theorem A.1; that source uses the vertex-count convention and $xy{x}^{-1}{y}^{-1}$ for individual group commutators.

Suppose $1\to X\to \mathcal{G}\to \Gamma \to 1$ is an exact sequence of pro-p groups, with $X$ abelian and $\Gamma \cong {\mathbb{Z}}_p$. Conjugation gives $X$ a compact Galois module structure. The closed commutator subgroup is $(\gamma -1)X$: its image is closed by compactness, and after quotienting by it, a lift of $\gamma$ centralizes $X$ and topologically generates the remaining quotient. Hence $0\to X_{\Gamma }\to {\mathcal{G}}^{\mathrm{ab}}\to \Gamma \to 0$ is exact.

An abelian quotient is a quotient group $G/N$ that is an abelian group. It exists only for a normal subgroup $N$, and it is abelian exactly when $N$ contains the commutator subgroup. Thus every abelian quotient factors through the abelianization $G/[G,G]$.

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.

If a nilpotent group is generated by $a_1,\dots ,a_d$, every element of its commutator subgroup is a product $[x_1,a_1]\cdots [x_d,a_d]$. The sets in the displayed formula need not be subgroups. Induct on the nilpotency class $c$. Collection in $G/{\gamma }_c(G)$ gives the desired product with a central error. Since ${\gamma }_c(G)=[{\gamma }_{c-1}(G),G]$, that error is a product $[y_1,a_1]\cdots [y_d,a_d]$ with $y_i\in {\gamma }_{c-1}(G)$. These factors are central, and ${\gamma }_{c-1}(G)$ centralizes $G\prime$, so replacing $x_i$ by $x_i{y}_i$ absorbs the error. For a finitely generated pro-p group, applying this formula in all finite quotients and using compactness makes its algebraically generated commutator subgroup closed.

For the symmetric group $S_3$, take the transpositions $x=(12)$ and $y=(23)$. Products of permutations apply the rightmost factor first. ThenThe last permutation sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$: two swaps fail to commute by an entire three-cycle.

Every group commutator has sign of a permutation $+1$, so $S_3^{\prime }\subseteq A_3$, the alternating group of even permutations. Conversely, the displayed group commutator generates $A_3=\{e,(123),(132)\}$. Therefore $S_3^{\prime }=A_3$. The two cosets are the even permutations and the odd permutations, and their multiplication is just multiplication of signs: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 $Z(S_3)=\{e\}$.

Put ${\mathcal{N}}_m=\{f:f(t)\equiv t(\mathrm{mod}{t}^{m+1})\}$ for $m\ge 1$. These are open normal subgroups and ${\mathcal{N}}_m/{\mathcal{N}}_{m+1}\cong ({\mathbb{F}}_p,+)$ via the coefficient of $t^{m+1}$. Thus $[\mathcal{N}:{\mathcal{N}}_m]=p^{m-1}$. The depth of a nonidentity series is ${\mathrm{ord}}_t(f(t)-t)-1$. If $f=t+a{t}^{r+1}+\cdots$ and $g=t+b{t}^{s+1}+\cdots$, expansion of substitution givesHere $[f,g]=f^{-1}\circ g^{-1}\circ f\circ g$. This formula controls commutator subgroups and normal subgroups.

For a solvable group, its derived length is the least nonnegative integer $r$ at which the derived series reaches the trivial group, with $G^{(0)}=G$ and $G^{(r+1)}$ the commutator subgroup of $G^{(r)}$. A nontrivial abelian group has derived length one; a nonabelian group with nontrivial abelian commutator subgroup has derived length two.

A finitely generated abelian group whose elements all have finite order is finite. Induct on derived length for a finitely generated soluble group with this property. Its abelianization is finite, so its commutator subgroup has finite index and is finitely generated by Schreier's lemma. That subgroup is soluble of smaller derived length and is still a torsion group, hence finite by induction. The ambient group is a finite extension of it and is finite.

A nilpotent group is finite exactly when its abelianization is finite; finite generation is not a separate hypothesis. Induct on the nilpotency class $c$. The abelian case is immediate. Otherwise $G/{\gamma }_c(G)$ has smaller class and the same abelianization, so is finite. Let $H=G$, $K={\gamma }_{c-1}(G)$. Their commutator subgroup $[H,K]={\gamma }_c(G)$ is central. Hence its commutator map is a homomorphism in each variable. Both centralizer quotients are finite because their kernels contain ${\gamma }_c(G)$. Finitely many pairs of coset representatives therefore generate the abelian group $[H,K]$, and its exponent divides the exponent of $H/C_H(K)$. It is finite, completing the induction.

A finite nonabelian simple group is a finite group that is a nonabelian group and a simple group. Every such group is a perfect group because its commutator subgroup is a nontrivial normal subgroup.

If a nilpotent group is generated by $a_1,\dots ,a_d$, every element of its commutator subgroup is a product $[x_1,a_1]\cdots [x_d,a_d]$. The sets in the displayed formula need not be subgroups. Induct on the nilpotency class $c$. Collection in $G/{\gamma }_c(G)$ gives the desired product with a central error. Since ${\gamma }_c(G)=[{\gamma }_{c-1}(G),G]$, that error is a product $[y_1,a_1]\cdots [y_d,a_d]$ with $y_i\in {\gamma }_{c-1}(G)$. These factors are central, and ${\gamma }_{c-1}(G)$ centralizes $G\prime$, so replacing $x_i$ by $x_i{y}_i$ absorbs the error. For a finitely generated pro-p group, applying this formula in all finite quotients and using compactness makes its algebraically generated commutator subgroup closed.

The group commutator measures the failure of two elements of a group to commute. Under the convention $[x,y]=x^{-1}{y}^{-1}xy$, it is the identity exactly when $xy=yx$. The commutator subgroup is generated by these elements. The alternative convention $xy{x}^{-1}{y}^{-1}$ generates the same commutator subgroup; individual formulas must specify their convention.

For $G=⟨a,t\mid t{a}^2{t}^{-1}=a^3⟩$, the surjective endomorphism $\theta (a)=a^2$, $\theta (t)=t$ has nontrivial kernel by Britton's lemma. The commutator subgroup is generated by $a_i=t^{i}a{t}^{-i}$ for $i\in \mathbb{Z}$. For $j>i$, its defining relations give $a_j^{2^{j-i}}=a_i^{3^{j-i}}$. Hence ${\theta }^k([a_i,a_j])=1$ whenever $k\ge j-i$. The union of the increasing kernels is normal, and the images of all $a_i$ commute in its quotient, proving the first inclusion. The second follows from finite residual contains surjective endomorphism kernels applied to every iterate.

Stable invertible matrices, with stabilization $g\mapsto \mathrm{diag}(g,1)$, form the stable general linear group. Its elementary subgroup is its commutator subgroup, and the displayed quotient is its abelianization. Elementary matrices lift along any surjective ring map, even when arbitrary invertible matrices do not lift. Over a field, determinant identifies $K_1$ with its multiplicative unit group.

Let $\pi :G\to G/[G,G]$ be the abelianization map and let $A$ be a finite $K$-approximate group. Suppose a coset progression $HP(x_1,\dots ,x_r;L_1,\dots ,L_r)$ lies in $\pi (A^4)$ and has size at least $\delta |\pi (A)|$. ThenA section of $\pi$ is multiplicative up to a bounded power of $A$ inside the commutator subgroup; successively separating the subgroup and progression coordinates proves the inclusion needed for this estimate.

A group is metabelian when its commutator subgroup is abelian, equivalently when its derived length is at most two.

A nontrivial minimal normal subgroup $N$ of a finite solvable group is abelian: its commutator subgroup is characteristic, and minimality plus solvability excludes $N^{\prime }=N$. A nontrivial Sylow subgroup of this abelian group is characteristic, so minimality makes $N$ a p-group. The characteristic subgroup of pth powers is proper, hence trivial by minimality. Therefore $N$ is an elementary abelian group.