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Let $p<q$ be distinct prime numbers. Every group of group order $p^2{q}^2$ is Abelian if and only if $p$ does not divide $q^2-1$.

For sufficiency, the Sylow theorems make the number of Sylow p-subgroups one of $1,q,q^2$, congruent to one modulo $p$. The divisibility hypothesis excludes both nontrivial choices, so that subgroup is normal. The number of Sylow subgroups for $q$ is one of $1,p,p^2$, congruent to one modulo $q$. The choice $p$ is impossible because $p<q$. If $q$ divided $p^2-1=(p-1)(p+1)$, primality and $q>p$ would force $q=p+1$, hence $(p,q)=(2,3)$; this pair violates the hypothesis. Thus the other Sylow subgroup is normal too. Their intersection is trivial, and their group commutators lie in that intersection, so the group is their direct product of groups. A group of prime-square order is abelian: its nontrivial center of a group is either the whole group or has cyclic quotient, and a cyclic central quotient also forces commutativity. Both factors are therefore abelian.

For necessity, suppose $p\mid q^2-1$. The multiplicative group of a finite field is cyclic, so ${\mathbb{F}}_{q^2}^{\times }$ contains $\alpha$ of order $p$. Let the cyclic group $C_{p^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by mapping its generator to multiplication by $\alpha$. This is a well-defined nontrivial group action, since ${\alpha }^{p^2}=1$. The resulting semidirect product has order $p^2{q}^2$ and is nonabelian, because the generator does not commute with a nonzero field element. Thus a nonabelian group of this order exists exactly when $p\mid q^2-1$.

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.

For $G_{\lambda }=\{g:\omega (g)\ge \lambda \}$ and $G_{\lambda +}=\{g:\omega (g)>\lambda \}$, the associated graded groupis a graded Lie algebra with bracket induced by the group commutator. For a p-valuation, $t{\mathrm{gr}}_{\lambda }(g)={\mathrm{gr}}_{\lambda +1}(g^p)$ makes it a graded ${\mathbb{F}}_p[t]$-Lie algebra.

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 $\left(\begin{array}{cc}i& 1\\ 2i& 2-i\end{array}\right)$.

Let $G$ be a nilpotent group of class at most $s$ in which every element has order at most $r$. A subgroup generated by $k$ elements has order at mostIndeed, collection by the lower central series expresses every element as an ordered product of simple group commutators in the generators of weights at most $s$. There are at most $k+k^2+\cdots +k^s\le s{k}^s$ such commutators, and each exponent may be reduced to one of at most $r$ values.

The commutator of two elements of an associative algebra over a field isIt measures how much the order of multiplication matters: $[A,B]=0$ exactly when $AB=BA$. It is a bilinear map, it is alternating since $AA-AA=0$, and associativity makes it satisfy the Jacobi identity. Hence every associative algebra becomes a Lie algebra on the same vector space when equipped with this product. This holds in every characteristic of a field.

For square matrices, this is the matrix commutator, and it is defined even for singular matrices. The group commutator of invertible matrices is instead $AB{A}^{-1}{B}^{-1}$: it equals the identity element when the matrices commute, whereas $AB-BA$ equals the zero matrix. The Lie bracket from local group commutators explains how differentiation connects the two constructions.

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

For odd $p$, group commutator collection in the depth filtration gives $[{\mathcal{N}}_m,{\mathcal{N}}_m]={\mathcal{N}}_{2m+1}$. Every degree $k\ge 2m+1$ occurs as a leading group commutator degree by choosing $r,s\ge m$, $r+s=k$, and $r\not\equiv s(\mathrm{mod}p)$; changing $r$ by one changes $r-s$ by two. Closed successive coefficient correction then fills the tail. Therefore ${\mathcal{N}}^{(n)}={\mathcal{N}}_{2^{n+1}-1}$ and ${\log }_p[\mathcal{N}:{\mathcal{N}}^{(n)}]=2^{n+1}-2$. In characteristic two the sharper collection theorem gives ${\log }_2[\mathcal{N}:{\mathcal{N}}^{(n)}]\le C{\kappa }^n$ for a constant $C$, with $\kappa =(1+\sqrt{17})/2<3$. In both cases the exponential base is strictly smaller than three.

For a smooth homomorphism of Lie groups, naturality gives $f(\exp X)=\exp (d{f}_{e}X)$. Therefore $\log (f(g))=d{f}_e(\log g)$ locally. Apply this to the group commutator and differentiate its two curve parameters to obtain the displayed identity for the Lie bracket from local group commutators. In particular the differential of a group representation is a Lie algebra representation.

Identify opposite sides of a square by translations to obtain a torus. All four corners become one vertex; the horizontal sides give one topological cell $a$ and the vertical sides one topological cell $b$. The open square supplies one two-dimensional topological cell. Thus $(c_0,c_1,c_2)=(1,2,1)$ and $\chi =0$. The cellular boundary of each one-dimensional topological cell is zero because its endpoints agree. Traversing the face boundary gives the group commutator $ab{a}^{-1}{b}^{-1}$, whose signed counts of both $a$ and $b$ vanish. Hence its cellular boundary is also zero, and $(b_0,b_1,b_2)=(1,2,1)$.^[ref]

The boundary of a tetrahedron is a two-sphere, with four vertices, six edges and four triangular topological cells: $\chi =4-6+4=2$. Its rational Betti numbers are $(1,0,1)$ by homology of a sphere. The torus's two independent one-dimensional homology classes cancel its degree-zero and degree-two contributions. The torus and annulus both have Euler characteristic zero, yet different homology groups; one integer cannot recover all of homology.

For a finite group, $F(G)$ centralizes its group socle. A nonabelian minimal normal subgroup meets $F(G)$ trivially: otherwise it is nilpotent, and its nontrivial characteristic center makes it abelian. Their group commutators therefore vanish. An abelian minimal normal subgroup $M$ is an elementary abelian p-group; $M\le O_p(G)$ and $M\cap Z(O_p(G))\ne 1$ by the fixed-point counting of a finite p-group on $M$. Minimal normality makes $M\le Z(O_p(G))$, and other p-cores commute with $M$.

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.

The Hall-Petrescu collection formula corrects the commutative rule $(xy{)}^n=x^n{y}^n$ by ordered products of higher group commutators. Its simplest useful form occurs in a two-step nilpotent group. With $[x,y]=x^{-1}{y}^{-1}xy$,Here $[y,x]$ is central. The relation $yx=xy[y,x]$ shows that moving the new $x$ past $n$ copies of $y$ adds $n$ to its exponent; mathematical induction and $\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)=\left(\genfrac{}{}{0ex}{}{n}{2}\right)+n$ prove the formula.

For an odd prime number $p$, the coefficient $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ is divisible by $p$. Thus if the commutator subgroup in this two-step setting has exponent dividing $p$, the $p$th power map satisfies $(xy{)}^p=x^p{y}^p$. For $p=2$ the correction generally remains. In higher nilpotency class, further commutators enter from deeper terms of the lower central series; their collection coefficients, rather than an assumption of commutativity, control power calculations in powerful p-groups and p-valued groups.

Use the group commutator convention $[x,y]=x^{-1}{y}^{-1}xy$ and write $u^v=v^{-1}uv$ for conjugation. The Hall-Witt identity isIt holds in every group. Substituting the definitions and cancelling adjacent inverse letters proves the formula directly; the three cyclic factors are all necessary for that cancellation.

Modulo the fourth term of the lower central series, the triple group commutators are central, so the outer conjugations disappear. Replacing an inverse input changes the corresponding leading group commutator by its inverse. In additive notation for the graded Lie algebra associated with the lower central series, the identity becomesthe Jacobi identity. This explains how a group-theoretic cancellation identity supplies the Lie-theoretic compatibility of leading commutators.

An internal direct product finds independent coordinates already present inside a group. Let $H,K$ be normal subgroups of $G$ withThen the multiplication mapis a group isomorphism. Thus every element of $G$ has a unique factorization into one element of each subgroup. Conversely, the embedded factors $G\times \{e_H\}$ and $\{e_G\}\times H$ inside an external direct product of groups satisfy these conditions.

For a concrete example, take the additive cyclic group $C_6=\mathbb{Z}/6\mathbb{Z}$ and its subgroups $H=\{0,3\}$ and $K=\{0,2,4\}$. They are normal subgroups because $C_6$ is an abelian group. Their intersection is $\{0\}$, and the sums $h+k$ give every residue class exactly once. For instance, $1=3+4\mathrm{mod}6$. The table makes the unique decomposition visible.

The proof explains each hypothesis. Normality puts the group commutator $[h,k]=h^{-1}{k}^{-1}hk$ in both $H$ and $K$, so it is the identity element. Therefore the factors commute, and $\Phi$ preserves the group operation. The condition $HK=G$ makes it surjective. If $hk=e$, then $h=k^{-1}$ belongs to the intersection, forcing both to be the identity element; thus its group kernel is trivial and it is injective.

Normality cannot simply be omitted. In the symmetric group $S_3$, take $H=⟨(123)⟩$ and $K=⟨(12)⟩$. Their intersection is trivial and $HK=S_3$, but $K$ is not a normal subgroup. With permutations composed right to left, $(123)(12)=(13)$ while $(12)(123)=(23)$, so the factors do not commute. The setwise decomposition is unique, but its multiplication map is not a group homomorphism from $H\times K$. An independent-coordinate interpretation requires the group operations to agree, not merely a bijection of the underlying sets.

In the setting of a uniform pro-p group $G$, its integral Lazard Lie algebra is a finite free ${\mathbb{Z}}_p$-module whose addition and Lie bracket recover infinitesimal versions of group multiplication and the group commutator. It carries more integral information than the associated ${\mathbb{Q}}_p$-Lie algebra. The construction requires hypotheses on the group and its filtration; it is not available unchanged for every profinite group.^[ref]

For odd $p$, a concrete example is $G=1+p{M}_n({\mathbb{Z}}_p)$. The convergent matrix logarithm and matrix exponential identify its Lie algebra with $p{M}_n({\mathbb{Z}}_p)$, using ordinary matrix addition and bracket $[A,B]=AB-BA$. Multiplication corresponds to the Baker--Campbell--Hausdorff formula, beginningFor $p=2$, the analogous safe neighborhood is $1+4M_n({\mathbb{Z}}_2)$ with Lie algebra $4M_n({\mathbb{Z}}_2)$. The smaller neighborhood $1+2{\mathbb{Z}}_2$ contains $-1$, so torsion already prevents it from being uniform.^[ref]

For the additive group ${\mathbb{Z}}_p$, the associated Lie algebra is rank one with zero bracket. Under suitable coefficient hypotheses, the Lazard isomorphism compares its Lie algebra cohomology with continuous group cohomology.

In a local exponential chart, the mixed second differential of the group commutator defines a bilinear map on the tangent space at the identity. Inversion after swapping the two group elements proves antisymmetry. Differentiating conjugation gives $[X,Y]=\mathrm{ad}(X)Y$; applying naturality of the Lie bracket to the Adjoint representation of a Lie group then proves the Jacobi identity. For a matrix group this recovers $XY-YX$ by expanding through the mixed term.