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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.

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 $G/[G,G]$.

With group commutator convention $[a,b]=a^{-1}{b}^{-1}ab$, the displayed group presentation describes the fundamental group of the Borromean rings complement. Its abelianization is ${\mathbb{Z}}^3$, since each relator is a group commutator and hence imposes no abelian relation. It embeds discretely in the Gaussian Bianchi group by sending the meridians to $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, $\left(\begin{array}{cc}1& 0\\ 2i& 1\end{array}\right)$ and $\left(\begin{array}{cc}i& 1\\ 2i& 2-i\end{array}\right)$.

The commutator subgroup of a group $G$ is the subgroup generated by all its group commutators:Here and throughout, $[x,y]=x^{-1}{y}^{-1}xy$. A group commutator is the identity element precisely when $x$ and $y$ commute. The generated subgroup includes finite products of group commutators and their inverse elements; an element of $G^{\prime }$ need not itself be a single group commutator. Thus $G^{\prime }=\{e\}$ precisely when $G$ is an abelian group.

The useful idea is to record exactly what must disappear to make multiplication commutative. The identity $xy=yx[x,y]$ shows how a group commutator corrects a change in multiplication order. Setting every such correction equal to $e$ produces the abelianization $G/G^{\prime }$, while preserving every piece of information that a group homomorphism to an abelian group can detect. This connects a local comparison of two operations to a canonical quotient group of the whole group.

The commutator subgroup is a characteristic subgroup, hence a normal subgroup: a group automorphism $\alpha$ sends $[x,y]$ to $[\alpha (x),\alpha (y)]$. For any normal subgroup $N$,Indeed, the group commutator of $xN$ and $yN$ is $[x,y]N$. This makes $G^{\prime }$ the smallest normal subgroup whose quotient group is an abelian group. See Conrad, Subgroup series I, section 4: commutators and abelian quotients.

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 $Z(S_3)=\{e\}$.

An endomorphism of $F_n$ has image generated by at most $n$ elements. If this image has finite index $d$, the Nielsen–Schreier formula gives rank $1+d(n-1)$. Its abelianization requires at least that many generators. For $n\ge 2$, comparison with $n$ forces $d=1$.

// Target: geometry-and-topology.bigb

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.

For a compact connected Riemannian manifold with nonnegative Ricci curvature, its finitely generated fundamental group acts freely and properly discontinuously on the complete universal cover. The preceding polynomial growth of properly discontinuous isometry groups bound applies. The degree-one Hurewicz theorem identifies first integral homology with the abelianization of the fundamental group. If its free rank is $b_1(M)$, this group maps onto ${\mathbb{Z}}^{b_1(M)}$. Choosing lifts of a basis gives at least a constant times $m^{b_1(M)}$ distinct elements in a word ball of radius a constant times $m$. The upper bound of order $m^{\dim M}$ forces the displayed inequality.

The first homology group $H_1(X)$ is the group of one-dimensional cycles modulo boundaries. For a path-connected space it is the abelianization of the fundamental group.

The integer Heisenberg group consists of upper-unitriangular three-by-three integer matrices. In coordinates $(x,y,z)$ its multiplication isIt is the semidirect product ${\mathbb{Z}}^2{⋊}_A\mathbb{Z}$ forand its abelianization is ${\mathbb{Z}}^2$.

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.

If $A$ is a finite $K$-approximate group in a torsion-free $s$-step nilpotent group, then there are $r\le K^{O(1)}$ approximate groups $A_0,\dots ,A_r\subseteq A^{O(1)}$, each with approximation parameter $K^{O(1)}$ and each generating a group of class less than $s$, such thatThe proof applies the large-progression form of the Freiman-Green-Ruzsa theorem in the abelianization, lifts its subgroup and cyclic directions, and uses the intersection of an approximate group power with a subgroup.

Every one-dimensional character kills the commutator subgroup and therefore factors through the abelianization. Conversely, every character of the abelianization pulls back to a one-dimensional character of the group.

The degree-sum formula for irreducible characters givesThe number of linear characters is the order of the abelianization, so $|G/G^{\prime }|=3$ and $|G^{\prime }|=7$. The intersection of the kernels of all linear characters is $G^{\prime }$. Consequently $g_4,g_5$ are nonidentity elements of this order-seven subgroup and have order seven. The images of $g_2,g_3$ under a displayed linear character have order three, so their orders are divisible by three. Lagrange's theorem leaves three or twenty-one; an element of order twenty-one would make $G$ cyclic, contradicting its degree-three irreducible characters. ThereforeThe center is trivial. A nontrivial proper center would have order three or seven, giving a cyclic quotient and hence an abelian group, again a contradiction. Thus a centralizer of a nonidentity element is proper. It has order three for an order-three element, and seven for an order-seven element. The class sizes are therefore $1,7,7,3,3$.

Column orthogonality says ${\sum }_i|{\chi }_i(g){|}^2=|C_G(g)|$. For $g_2,g_3$, the three linear characters already contribute three, so the remaining two entries must both vanish. Orthogonality with the identity column says ${\sum }_i{\chi }_i(1)\overline{{\chi }_i(g)}=0$ for $g\ne 1$. At $g_4$ this gives $1+\alpha +{\chi }_5(g_4)=0$, so ${\chi }_5(g_4)=\beta$, using $\alpha +\beta =-1$. Similarly ${\chi }_5(g_5)=\alpha$. Thus the completed table isHere $\beta =\overline{\alpha }$, $|\alpha {|}^2=|\beta {|}^2=2$ and $\alpha \beta =2$. These relations, together with $1+\omega +{\omega }^2=0$, verify all weighted row orthogonality relations and the centralizer column norms.

Suppose inductively that $G=H{\gamma }_i(G)$. In the quotient $G/{\gamma }_{i+1}(G)$, the image of ${\gamma }_i(G)$ is central. Hence this quotient is generated by the image of $H$ together with a central subgroup, so its commutator subgroup lies in the image of $H$. Pulling this back givesThe original assumption now gives $G=H{\gamma }_2(G)\le H{\gamma }_{i+1}(G)\le G$. Starting with $i=2$, induction proves $G=H{\gamma }_i(G)$ for every $i\ge 2$. These products are subgroups because each ${\gamma }_i(G)$ is normal; no normality of $H$ is needed.

If $G$ is finitely generated, every quotient, including its abelianization $G/{\gamma }_2(G)$, is finitely generated. Conversely, lift a finite generating set of that quotient to $g_1,\dots ,g_t\in G$, and set $H=⟨g_1,\dots ,g_t⟩$. Then $G=H{\gamma }_2(G)$. For a nilpotent group, ${\gamma }_{c+1}(G)=1$ for some $c$, and the established identity gives $G=H$. Thus

All commutators of $H$ with $K$ lie in the central subgroup $Y$ of $HK$. In the standard commutator identities, conjugation by elements of $H$ or $K$ therefore acts trivially on these commutators. Since $Y$ is abelian, their order can also be reversed. ConsequentlyThus the map is a homomorphism in each variable separately; this does not assert that it is a homomorphism from the direct product jointly.

It factors through $H/C_H(K)$ in the first variable and $K/C_K(H)$ in the second. If these finite quotient groups have representatives $h_1,\dots ,h_a$ and $k_1,\dots ,k_b$, then the finite collection $[h_i,k_j]$ generates $Y$. Let $e$ be an exponent of $H/C_H(K)$. For every $h,k$,Hence $Y$ is a finitely generated abelian group of bounded exponent, and $Y$ is finite. For instance its order is at most $e^{ab}$.

Now suppose $G$ is nilpotent and $G/{\gamma }_2(G)$ is finite. Induct on the nilpotency class $c$. For $c=1$, $G$ is its own abelianization and is finite. If $c\ge 2$, the group $\bar{G}=G/{\gamma }_c(G)$ has smaller class andsince ${\gamma }_c(G)\le {\gamma }_2(G)$. By induction $\bar{G}$ is finite. Apply the preceding result with $H=G={\gamma }_1(G)$ and $K={\gamma }_{c-1}(G)$. Their commutator subgroup is $Y={\gamma }_c(G)\le Z(G)$. Both relevant centralizer quotients are finite, because their centralizers contain ${\gamma }_c(G)$ and $G/{\gamma }_c(G)$ is finite. Thus ${\gamma }_c(G)$ is finite. The extension of this finite normal subgroup by the finite quotient $\bar{G}$ is finite. The converse follows because a quotient of a finite group is finite. Therefore