Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 7 (Conjugacy, the center and the five-eighths bound)

Write $g\cdot x=gx{g}^{-1}$. If $e$ is the identity element of $G$, then $e\cdot x=ex{e}^{-1}=x$. Also, for $g,h,x\in G$,The middle equality uses associativity and the inverse element rule $(gh{)}^{-1}=h^{-1}{g}^{-1}$. Thus this is a left group action, the conjugation action of $G$ on itself.

The orbit of $x$ under conjugation is its conjugacy classIt contains $x$ by using $g=e$, so the conjugacy classes cover $G$. If $z=gx{g}^{-1}$, then $x=g^{-1}zg$, and the group action identity shows that conjugating either element further gives the same orbit; hence $C(z)=C(x)$. If $C(x)$ and $C(y)$ intersect at $z$, this gives $C(x)=C(z)=C(y)$. Thus distinct conjugacy classes are disjoint and the conjugacy classes form a partition of a set $G$.

The specified set $A$ is the center of a group $Z(G)$. It contains $e$. If $a,b\in A$ and $y\in G$, thenso $ab\in A$. From $ay=ya$, multiplying on the left and right by $a^{-1}$ gives $y{a}^{-1}=a^{-1}y$, so $a^{-1}\in A$. Hence $A$ is a subgroup.

For $g\in G$ and $a\in A$, the defining commutativity gives $ga{g}^{-1}=ag{g}^{-1}=a$. Thus the image of each element of $A$ under conjugation is in $A$, and indeed $gA{g}^{-1}=A$. Consequently $A=Z(G)$ is a normal subgroup of $G$.

Suppose the quotient group $G/A$ is a cyclic group, generated by $gA$. For any $x,y\in G$, there are integers $r,s$ and $a,b\in A$ such thatSince $a,b$ lie in the center of a group, they commute with every element, in particular with each other and with the integer powers of a group element $g$. ThereforeAs $x,y$ were arbitrary, $G$ is an abelian group. This is the cyclic quotient by the center.

A smallest example is the Klein four-groupwith coordinatewise addition modulo $2$. Every nonidentity element has order of a group element $2$, so none generates all four elements; hence this group is not a cyclic group.

A group of group order $1$ is a cyclic group. A group of group order $2$ or $3$ is also a cyclic group: choose a nonidentity element; by Lagrange's theorem, its order of a group element is a nontrivial divisor of that prime number, so equals the whole group order. Thus no smaller example exists, and the smallest possible group order is $4$.

If $G$ is not Abelian, the contrapositive of ii says that $G/A$ is not a cyclic group. The quotient group is finite because $G$ is finite. By iii, every group with fewer than four elements is a cyclic group. HenceEquivalently, the center of a group satisfies $|A|\le |G|/4$, by the coset count in iii.

A conjugacy class $C(x)$ is a singleton exactly when $gx{g}^{-1}=x$ for every $g\in G$, equivalently when $gx=xg$ for every $g$. Thus the singleton conjugacy classes correspond exactly to elements of the center of a group $A$, and there are $|A|$ of them.

Every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, if $r$ is the number of these remaining classes then $2r\le |G|-|A|$. ConsequentlyThis conjugacy class count bound from the center needs no classification of the finite group.

For a nonabelian group, iv and iii imply $|G|=|G/A||A|\ge 4|A|$. Substituting $|A|\le |G|/4$ into i yieldsThe constant is sharp: the quaternion group has eight elements and the five conjugacy classes $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$, $\{k,-k\}$. To verify this directly, $\pm 1$ are central elements, conjugation by $j$ sends $i$ to $-i$, and similarly for $j,k$; its relations $i^2=j^2=k^2=-1$ and $ij=k=-ji$, $jk=i=-kj$, $ki=j=-ik$ show that conjugating $i,j,k$ by any of $\pm 1,\pm i,\pm j,\pm k$ changes only its sign. Hence $m(G)=5=(5/8)|G|$.