Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 8D (Groups)

The identity belongs to the center of a group. If $z,w$ commute with every $g$, then $zw$ does too; rearranging $gz=zg$ also shows that $z^{-1}$ commutes with $g$. Hence $Z(G)$ is a subgroup. Moreover $gz{g}^{-1}=z$ for $z\in Z(G)$, so it is a normal subgroup.

Suppose $|G|=p^n>1$. The conjugation action partitions $G$ into conjugacy classes. The orbit-stabilizer theorem gives class size $[G:C_G(x)]$, where $C_G(x)$ is the centralizer of $x$. By Lagrange's theorem, this size is a power of $p$; it is one exactly when $x$ is central. Thus the class equation has the formIt follows that $p\mid |Z(G)|$. Since the identity is central, $|Z(G)|\ge 1$, and thereforeThis is the nontrivial-centre property of a finite p-group.

Write the Heisenberg group element as $M(x,y,z)$. Matrix multiplication and inversion giveAll matrices have determinant one. They contain $M(0,0,0)=I$, are closed under multiplication and inverses, and inherit associativity from matrix multiplication; hence they form a subgroup of $G{L}_3(\mathbb{R})$.

Two such matrices commute precisely when $x{z}^{\prime }=x^{\prime }z$. Requiring this for every $x^{\prime },z^{\prime }$ forces $x=z=0$. ConsequentlyThe map $\pi (M(x,y,z))=(x,z)$ is a surjective group homomorphism to the additive group ${\mathbb{R}}^2$, with kernel of a group homomorphism $Z(H)$. Equivalently, it gives the explicit quotient group isomorphismIt is well-defined because multiplying by a central element changes only $y$. It is injective because equal $(x,z)$ imply that the quotient of the two matrices is central, and surjectivity follows by taking $y=0$. This also proves the needed instance of the first isomorphism theorem directly.

For the modular Heisenberg group, the three independent entries each have $p$ choices, soThe multiplication and commutation calculations work over the finite field ${\mathbb{F}}_p$ exactly as over $\mathbb{R}$. Taking $x^{\prime }=1,z^{\prime }=0$ and $x^{\prime }=0,z^{\prime }=1$ shows thatLet $c=M(0,1,0)$. Its powers exhaust the centre and $c$ has order $p$, giving the group presentationLet $a=M(1,0,0)$ and $b=M(0,0,1)$. Their images $\bar{a},\bar{b}$ in the quotient group translate the $x,z$ entries independently. They commute modulo the centre, have order $p$, and generate $p^2$ distinct pairs. ThereforeFor context, $ab{a}^{-1}{b}^{-1}=c$, with $c$ central. All conclusions above include $p=2$; one must not infer that every nonidentity element of $H_2$ has order two, since $(ab{)}^2=c$.