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The identity commutes with every group element, so it lies in the center of a group. If $z,w$ are central and $g\in G$, then $gw=wg$ implies $g{w}^{-1}=w^{-1}g$. ConsequentlyThus $z{w}^{-1}$ is central. The subgroup test provesEquivalently, the centre is nonempty, closed under products, and closed under inverses. This proof does not require finiteness. In fact the centre is an Abelian subgroup, and it is normal because $gz{g}^{-1}=z$ for each central element.

Denote a matrix by $A(x,y,z)$, with $y$ the upper-right entry. Direct matrix multiplication givesThis proves closure. Associativity is inherited from matrix multiplication. The identity is $A(0,0,0)$, and the inverse isBoth products with this matrix are the identity. Thus these real matrices form the Heisenberg group, in the coordinate convention with central coordinate $y$.

Let $N=\{A(0,b,0):b\in \mathbb{R}\}$. Its multiplication adds $b$, its identity is $b=0$ and its inverse negates $b$, so it is a subgroup. MoreoverThus every element of $N$ is central, and $gN{g}^{-1}=N$. Therefore $N$ is a normal subgroup. In fact it is the full center of a group, as described by the center quotient of the real Heisenberg group.

The center of a group contains $e$. If $z,w$ lie in the center, then for every $g\in G$,It is therefore closed under the group operation and inverse elements, and inherits associativity, proving it is a subgroup. Each of its elements commutes with every element of $G$, in particular with every other central element. HenceThus the center is an abelian subgroup.

Factor the group order as $588=2^2\cdot 3\cdot 7^2$. The Sylow theorems say that the number of Sylow subgroups for the prime seven divides $12$ and is congruent to one modulo seven. Among $1,2,3,4,6,12$, only one qualifies. Thus there is a unique normal subgroup $N$ of order $49$.

Every group of prime-square order is Abelian. Indeed its center of a group is nontrivial, and a cyclic quotient by the center of a group forces the whole group to be abelian. Consequently $N$ is either $C_{49}$ or $C_7\times C_7$, and in either case it has two composition factors, both $C_7$.

It remains to consider $Q=H/N$, of order twelve. Its number of Sylow subgroups for the prime three is either one or four:

Lift the composition series of $Q$ to $H$ using the subgroup correspondence theorem, and join it to a composition series inside $N$. This givesas the multiset of composition factors. The Jordan–Hölder theorem makes this independent of the chosen series. In particular every group of order $588$ is a solvable group, with composition length five; that is a statement about composition length, not necessarily its derived length.

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.

Let $|G|=p^2$. The allowed center of a group theorem gives $|Z(G)|>1$. By Lagrange's theorem its order is $p$ or $p^2$. In the first case $G/Z(G)$ has order $p$ and is a cyclic group. A cyclic quotient by the center forces $G$ to be Abelian: if $G/Z(G)=⟨gZ(G)⟩$, write arbitrary elements as $g^{a}z,g^{b}w$ with $z,w\in Z(G)$, and commute them directly. Hence in every caseFor two examples at prime-cube order, take the dihedral group of a square, $⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, and the quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$. Both have order $8=2^3$. They are nonabelian because $sr\ne rs$ and $ij=k\ne -k=ji$, respectively. They are distinct examples even up to group isomorphism: the dihedral group has five elements of order two and the quaternion group has only one. The request asks for two examples; it does not require the prime to be arbitrary.

Let $G$ be the real upper unitriangular three-by-three group, let $Z$ be its centre, and let $Z_0$ be the integer subgroup of $Z$. Every finite-dimensional continuous complex representation of $G/Z_0$ kills the central circle $Z/Z_0$, and hence is not faithful.

On $L^2(\mathbb{R})$, translations and multiplication by phases give the Schrödinger representation of the Heisenberg group. The central coordinate acts by an overall phase.

A group $G$ is two-step nilpotent whenEquivalently, its commutator subgroup is contained in the center of a group of $G$. This allows every word to be collected into powers of generators followed by powers of their pairwise commutators.

Repeatedly take the center of a group after quotienting by the preceding term. A group is a nilpotent group exactly when a finite term is the whole group. This yields the normalizer condition for nilpotent groups.