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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 $p\mid q^2-1$.

If an abelian subgroup $A$ has index $p^a$, the centralizer of any $1\ne g\in A$ contains $A$, so the class size of $g$ divides $p^a$. The prime-power conjugacy-class obstruction to simplicity rules this out. If $A=1$, the group is a p-group and has nontrivial center of a group, also ruling out nonabelian simplicity.

Like the center of a group, the Lie-algebra center consists of elements commuting with every element; here commutation means vanishing of the Lie bracket.

The center is $Z(\mathfrak{g})=\{z:[z,x]=0\text{ for every }x\in \mathfrak{g}\}$.

In coordinates $h(x,y,z)=\left(\begin{array}{ccc}1& x& y\\ 0& 1& z\\ 0& 0& 1\end{array}\right)$, the Heisenberg group law is $h(x,y,z)h(a,b,c)=h(x+a,y+b+xc,z+c)$. Commutation for every $(a,b,c)$ forces $xc=az$, hence $x=z=0$. The center of a group is therefore the $y$-axis. The surjective group homomorphism $h(x,y,z)\mapsto (x,z)$ has exactly this kernel of a group homomorphism, giving the stated quotient group by the first isomorphism theorem.

An element of a group is central when it commutes with every element of that group. The set of all such elements is the center of a group. Equivalently, its conjugacy class is a singleton: $xg{x}^{-1}=g$ for every $x$ is exactly the commutation condition. This lets centrality identify one-element conjugacy classes without enumerating every conjugation.

A nontrivial normal subgroup $N$ of a finite p-group meets the center of a group nontrivially. The conjugation action of $G$ on $N$ has singleton orbits exactly at $N\cap Z(G)$. The orbit-stabilizer theorem makes every other orbit size divisible by $p$, so $|N\cap Z(G)|\equiv |N|\equiv 0(\mathrm{mod}p)$. Since the intersection contains the identity, its size is at least $p$.

The centralizer of an element $x$ of a group $G$ consists of all elements that commute with $x$:It is the stabilizer subgroup of $x$ under the conjugation action, hence a subgroup of $G$. In conjugacy classes of the symmetric group on three letters, the three representative computations areFor the transposition $(12)$, a commuting permutation must preserve the exchanged pair $\{1,2\}$ and fix $3$; only $e$ and $(12)$ do so. For $r=(123)$, a commuting permutation is determined by where it sends $1$: the equation $g(r(i))=r(g(i))$ forces the images of $2$ and $3$, leaving exactly $e,r,r^2$. For $e$, every permutation commutes. These centralizer sizes $6,2,3$ give conjugacy class sizes $1,3,2$.

The centralizer is useful because it measures how much freedom remains after fixing one operation. Its index of a subgroup counts the distinct results of conjugation, and the intersection of all element centralizers is the center of a group. A centralizer need not be an abelian group: $C_G(e)=G$ even when $G$ is a nonabelian group. In an abelian group, all element centralizers equal $G$. These observations also explain why the surrounding group matters: the centralizer of $(123)$ inside the alternating group $A_3$ is the whole $A_3$, whereas its centralizer inside $S_3$ is a proper subgroup. See McKernan, the centralizer definition preceding lemma 23.6.

For an arbitrary subset $A\subseteq G$, the centralizer of a subset fixes every element of $A$ individually under the conjugation action:This is a pointwise stabilizer of the elements of $A$, not of the vertices of a polygon in a geometric realization. In particular, $r=(123)$ belongs to $C_{S_3}(r)$ although its rotation moves every vertex of a polygon of the equilateral triangle. If $A$ generates a subgroup $H$, commuting with every element of $A$ implies commuting with every product of those elements and their inverse elements; hence $C_G(A)=C_G(H)$. For the triangle rotation subgroup $H=⟨r⟩$, this gives $C_{S_3}(H)=C_{S_3}(r)=H$.

To see the commutation test directly on the equilateral triangle, compare a rotation $r$ with the reflection $s=(12)$. Their permutations satisfyThe two orders send vertex of a polygon $1$ to different places: $rs(1)=3$, but $sr(1)=1$. Thus $s\notin C_{S_3}(r)$. By contrast $r$ and $r^2$ commute, with $r{r}^2=r^{2}r=e$. The reflection reverses the chosen rotation while preserving the subgroup of all triangle rotations, so $s\in N_{S_3}(H)$. The distinction is visible before any counting: commuting preserves an operation itself; membership in the normalizer permits exchanging it with another operation in the same subgroup.

For a finite group $G$, choose one representative $x_i$ from each conjugacy class with more than one element. The class equation iswhere $Z(G)$ is the center of a group and $C_G(x_i)$ is the centralizer of $x_i$. Each central element contributes its own one-element conjugacy class. Every remaining conjugacy class contributes its size, given by the orbit-stabilizer theorem. The sum counts each element exactly once. See McKernan, Group actions and automorphisms, lemma 23.6.

For conjugacy classes of the symmetric group on three letters, take $x_1=(12)$ and $x_2=(123)$. Their centralizers have sizes two and three, and the center of a group has size one, givingFor an abelian group, every centralizer is the whole group, every conjugacy class has size one, and the sum over noncentral representatives is empty.

The power of this counting identity is that the sizes of conjugacy classes constrain group structure. For a nontrivial finite p-group with group order $p^n$, every noncentral conjugacy class has size a positive power of $p$, by Lagrange's theorem. The class equation therefore gives $|Z(G)|\equiv 0(\mathrm{mod}p)$. Since the identity element lies in the center of a group, the center of a group must have at least $p$ elements. This obtains an element commuting with the entire group without constructing one. See McKernan, lemma 23.7.

Historically, these ideas belong to the nineteenth-century development of permutation groups and normal subgroups. Milne's historical opening records Galois's introduction of normal subgroups in 1832 and Jordan's systematic treatment of permutation groups in 1870. This is context for the conjugation action and the class equation, rather than a claim that either author first stated the modern formula.^[ref]

A classical group is a standard family of linear groups, principally the general/special linear groups and groups preserving nondegenerate alternating, symmetric, Hermitian or quadratic forms. Finite-field versions yield many finite simple groups after imposing the appropriate determinant or derived-subgroup condition and quotienting the scalar center of a group. Small dimensions and small fields have exceptional isomorphisms and nonsimple cases.

The nontrivial center of a finite p-group theorem implies that the center of a group of order $p^2$ has order $p$ or $p^2$. If it has order $p$, the quotient by the center has prime order and is cyclic; a cyclic quotient by the center forces the whole group to be abelian, contradicting the assumed size of its center. Thus the group is abelian. An element of order $p^2$ makes it cyclic. Otherwise choose $a\ne 1$ and $b\notin ⟨a⟩$; their order-$p$ cyclic subgroups intersect trivially and generate the direct product of groups $C_p\times C_p$. The two possibilities are nonisomorphic because only the cyclic one has an element of order $p^2$.

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 an element $x$ of a group $G$, its conjugacy class is its orbit of a group action under the conjugation action:Thus conjugate group elements describe the same operation after a change of labels. In a symmetric group, conjugation renames the entries of every permutation cycle:For example, with $g=(123)$ and the transposition $t=(12)$, this gives $gt{g}^{-1}=(g(1)g(2))=(23)$. Products use function composition of permutations, with the rightmost factor applied first. The complete six-element example is conjugacy classes of the symmetric group on three letters.

The conjugacy classes form a partition of a set of $G$, since conjugacy is an equivalence relation. In a symmetric group, two permutations lie in the same conjugacy class precisely when they have the same cycle type: renaming preserves the lengths of permutation cycles, and matching equal-length permutation cycles constructs a renaming in the other direction. This compresses many labelled permutations into a few structural types. See McKernan, Conjugation in Sn, lemma 6.2 and proposition 6.6.

The stabilizer subgroup of $x$ under the conjugation action is its centralizer $C_G(x)$. Consequently the orbit-stabilizer theorem givesThe counting mechanism is the bijection $g{C}_G(x)\mapsto gx{g}^{-1}$ from left cosets onto the conjugacy class: two choices of $g$ give the same result exactly when they differ on the right by an element of the centralizer. A large centralizer means many choices of labels leave the operation unchanged, hence a small conjugacy class. A conjugacy class has size one precisely when its element belongs to the center of a group. The class equation counts these orbits of a group action together.

For a finite group $G$, write $k(G)$ for its number of conjugacy classes. Exactly the elements of the center of a group $Z(G)$ have singleton conjugacy classes; every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, $k(G)-|Z(G)|\le (|G|-|Z(G)|)/2$. If $G$ is a nonabelian group, the cyclic quotient by the center criterion and the fact that every group of group order at most three is a cyclic group give $[G:Z(G)]\ge 4$, hence $k(G)\le 5|G|/8$. The quaternion group attains this bound, with five conjugacy classes among eight elements.

The symmetric group $S_3$ has exactly six permutations, written using permutation cycles:Here $e$ is the identity element; $(12)$, $(13)$ and $(23)$ are transpositions; and $(123)$ and $(132)$ are permutation cycles of length three. With $t=(12)$ and $r=(123)$, rightmost-first function composition gives $r^2=(132)$ and $r^{-1}=r^2$. The conjugacy classes and representative centralizers are

An explicit check of the conjugation $gt{g}^{-1}$ with $g=r=(123)$ sends $1\mapsto 3\mapsto 3\mapsto 1$, $2\mapsto 1\mapsto 2\mapsto 3$, and $3\mapsto 2\mapsto 1\mapsto 2$, where each chain applies $g^{-1}$, then $t$, then $g$. Thus $1$ is fixed and $2,3$ are exchanged: $gt{g}^{-1}=(23)$. Relabelling also produces $(13)$, so the three transpositions share one conjugacy class. A transposition reverses the length-three permutation cycle, since $tr{t}^{-1}=(213)=(132)=r^{-1}$, putting both length-three permutation cycles in one conjugacy class.

Checking each of the six possible permutations against $gx=xg$ gives the displayed centralizers. Only $e$ has a one-element conjugacy class, so the center of a group here is $Z(S_3)=\{e\}$. The class equation is $6=1+3+2$. The second diagram in centralizer and normalizer uses the same $r$ and $t$ to show how preserving a cyclic subgroup differs from fixing its group generator.

Induct on the order of the finite p-group. Its center of a group is nontrivial unless the group itself is trivial. Quotient by the center to obtain a smaller finite p-group and lift a central series supplied by induction. Prepending the center gives a central series for the original group: the first term is central, and the remaining commutator-containment conditions are exactly the lifted quotient conditions. Hence every finite p-group is a nilpotent group. Since central factors are Abelian, every nilpotent group is also a soluble group.

These are all the groups of group order $70$ up to group isomorphism, with $D_n$ denoting a dihedral group of order $n$. The Sylow theorems make the Sylow subgroups of orders five and seven unique: the possible counts divide fourteen and ten, respectively, and only one has the required residue. These normal subgroups of coprime order commute, giving a normal subgroup $N\cong C_{35}$. Cauchy's theorem for finite groups provides an element of order two outside $N$, so $G=N⋊C_2$.

The automorphism of a finite cyclic group defining this semidirect product is a unit $a$ modulo $35$ with $a^2\equiv 1$. The Chinese remainder theorem makes its restrictions modulo five and seven independently $+1$ or $-1$. The resulting exponents are $1,34,6,29$, giving the four displayed groups in order. Their center of a group orders are $70,1,5,7$, so no two are isomorphic. Every case exists as the specified semidirect product.

For a commutative ring $R$ with an identity element, use coordinatesThese matrices form a group under multiplication, withThese formulas require no division, so they remain valid over rings with zero divisors. The center of a group is $\{A(0,y,0):y\in R\}$: the commutator has only central coordinate $x{z}^{\prime }-x^{\prime }z$, and testing $x^{\prime }=0,z^{\prime }=1$ and $x^{\prime }=1,z^{\prime }=0$ forces $x=z=0$. For finite $R$, its group order is $|R{|}^3$. This includes real, integer and modular versions of the Heisenberg group; the ring $\mathbb{Z}/4\mathbb{Z}$ must not be confused with the finite field of four elements.

The modular Heisenberg group consists of upper-unitriangular $3\times 3$ matrices over the finite field ${\mathbb{F}}_p$. In coordinates $M(x,y,z)$, multiplication is $M(x,y,z)M(x\prime ,y\prime ,z\prime )=M(x+x\prime ,y+y\prime +xz\prime ,z+z\prime )$, and its order is $p^3$. Its center of a group is $\{M(0,y,0)\}\cong C_p$; the quotient group by the centre is $C_p\times C_p$, represented by the $x,z$ coordinates. The elementary generators $a=M(1,0,0)$, $b=M(0,0,1)$, $c=M(0,1,0)$ satisfy $a^p=b^p=c^p=e$, $ab{a}^{-1}{b}^{-1}=c$, and $c$ central. For $p=2$, $(ab{)}^2=c$, so the whole group does not have exponent two.

Every nontrivial finite p-group has a nontrivial center. The conjugation action partitions the group into conjugacy classes whose noncentral sizes are positive powers of $p$; the class equation then implies $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$.

For a subgroup $H$ of a group $G$, the normalizer isIt is the stabilizer subgroup of $H$ under the conjugation action on subgroups. Its elements can permute members of $H$; they need not commute with them. For $r=(123)$ in the symmetric group $S_3$, the cyclic subgroup $H=⟨r⟩$ has normalizer $S_3$, although the centralizer of $r$ has only three elements, as illustrated in centralizer and normalizer.

Every element of $H$ preserves $H$ under conjugation, so $H\subseteq N_G(H)$. Moreover, $H$ is a normal subgroup of $N_G(H)$, and this is the largest subgroup of $G$ with that property. Indeed, any subgroup in which $H$ is a normal subgroup consists entirely of elements preserving $H$ under conjugation. In particular, $N_G(H)=G$ exactly when $H$ is a normal subgroup of $G$.^[ref]

The orbit-stabilizer theorem counts distinct conjugate subsets of a subgroup byFor $H=⟨(123)⟩$ in $S_3$, there is just one, since $[S_3:N_{S_3}(H)]=6/6=1$. For the subgroup $K=⟨(12)⟩=\{e,(12)\}$, preserving $K$ forces its unique nonidentity element to stay fixed. Hence $N_{S_3}(K)=C_{S_3}((12))=K$, and its three distinct conjugate subsets are $\{e,(12)\}$, $\{e,(13)\}$ and $\{e,(23)\}$. This is the subgroup counterpart of counting a conjugacy class with a centralizer: the object being stabilized determines which denominator belongs in the count.^[ref]

The equality in the definition is essential. For a subgroup $H$ that is a finite group, an inclusion $gH{g}^{-1}\subseteq H$ already gives equality by equal cardinalities. For an infinite subgroup, that inference can fail: conjugation might carry it onto a proper subgroup. The normal subgroup criterion is therefore, with the explicit hypothesis $H\le G$,No finiteness or commutativity assumption is needed. A normal subgroup is preserved as a set; its elements need not lie in the center of a group.

There is a second useful count, distinct from counting conjugate subgroups. The normalizer acts on $H$ itself by group automorphisms, giving the group homomorphismIts group kernel is $C_G(H)$, so the centralizer of a subset is a normal subgroup of $N_G(H)$ and the first isomorphism theorem givesThis quotient group records the different rearrangements of $H$ induced by the surrounding group. It gives a faithful group action on $H$, but need not be the entire automorphism group of $H$. For example, take the ambient group $G=H=⟨(123)⟩$. Since $G$ is an abelian group, $N_G(H)=C_G(H)=G$, so the quotient group is trivial; nevertheless $r\mapsto r^{-1}$ is a nonidentity group automorphism of $H$. Returning to the ambient symmetric group $S_3$ makes that group automorphism available through a reflection, and $N_{S_3}(H)/C_{S_3}(H)\cong C_2$. Its two cosets are the three rotations and the three reflections.

The subgroup $H$ itself need not be contained in $C_G(H)$: their intersection is the center of a group $Z(H)$. It is contained precisely when $H$ is an abelian group. Thus $N_G(H)/H$, which exists because $H$ is a normal subgroup of its normalizer, and $N_G(H)/C_G(H)$ answer different questions and should not be substituted for one another.