Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 6

A subgroup $N$ of $G$ is a normal subgroup if conjugation by every element preserves it:Equivalently, $gN=Ng$ for every $g\in G$. Here equality means equality of sets; it does not require each individual element of $N$ to commute with every element of $G$.

The center of a group consists of its elements commuting with every element. First check it is a subgroup. The identity element is central. If $a,b\in Z(G)$ and $x\in G$, then $b^{-1}$ commutes with $x$ andThus $a{b}^{-1}\in Z(G)$, proving closure under the subgroup criterion. Moreover, if $z\in Z(G)$, then $xz{x}^{-1}=z$ for every $x\in G$. Hence conjugation fixes the center elementwise, soTherefore the center of a group is a normal subgroup.

If $g\in Z(G)$, it commutes with each $x\in G$, so $x^{-1}gx=g$. Its conjugacy class is therefore $C(g)=\{g\}$.

Conversely, if $C(g)=\{g\}$, then $x^{-1}gx=g$ for every $x$. Multiplication by $x$ on the left gives $gx=xg$ for every $x$, which is exactly membership in the center of a group. Thus

Conjugation of a permutation relabels each symbol by the inverse conjugating permutation. Specifically,To verify the direction of the inverse, apply the left side to ${\tau }^{-1}(a_j)$: $\tau$ first takes it to $a_j$, the original cycle takes it to $a_{j+1}$, and ${\tau }^{-1}$ takes it to ${\tau }^{-1}(a_{j+1})$, with cyclic indexing. Any symbols fixed by the original permutation are similarly relabelled and remain fixed. More generally, conjugate every cycle of a disjoint-cycle decomposition in this way; disjointness is preserved because ${\tau }^{-1}$ is a bijection.

By the preceding relabelling formula, every conjugate of $(123)$ is a three-cycle, and any three-cycle on four symbols can be obtained by a suitable relabelling. ThereforeThere are eight elements, agreeing with the count $\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot 2$.

Likewise, conjugation preserves the pattern of two disjoint transpositions, and every partition of four symbols into two pairs is realized by relabelling. ThusThese have the respective cycle types $(3,1)$ and $(2,2)$ in the symmetric group $S_4$.

Suppose $\sigma \ne e$. Choose a symbol $i$ moved by $\sigma$ and write $j=\sigma (i)\ne i$. Since $n\ge 3$, choose a third symbol $k$ distinct from $i,j$. Let $\tau =(jk)$, a transposition that fixes $i$. ThenThus ${\tau }^{-1}\sigma \tau \ne \sigma$, proving that the conjugacy class of any nonidentity permutation has more than one element. By part (a)(iii), no such permutation lies in the center of a group. The identity always does, soThe restriction on $n$ is necessary: $S_2$ is an abelian group and equals its own center.