Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 3 / 5D / Solution

For the centralizer of a subset, the identity commutes with every $h\in A$. If $g,k$ commute with each member of $A$, then for every such $h$,Also $gh=hg$ implies $g^{-1}h=h{g}^{-1}$ after multiplication by $g^{-1}$ on both sides. Thus identity, products and inverses lie in $C(A)$, proving that it is a subgroup.

For the conjugation action, identity acts trivially, andThese are the defining laws of a left group action. Its orbits of a group action are the conjugacy classes. The stabilizer of $h$ is exactly its centralizer, since $gh{g}^{-1}=h$ if and only if $gh=hg$. The orbit-stabilizer theorem therefore givesA class has one element exactly when that element commutes with all of $G$, so the singleton classes are precisely the elements of $C(G)$, the centre of a group. Partitioning $G$ into all its conjugacy classes now proves the class equationFor $|G|=p^r$, Lagrange's theorem for finite groups makes every class size a power of $p$. A nonsingleton class has size at least $p$ and is therefore divisible by $p$. Reducing the class equation modulo $p$ yields $|C(G)|\equiv 0(\mathrm{mod}p)$, because $r>0$. The identity lies in $C(G)$, so its order is positive and divisible by $p$. ConsequentlyThis proves the nontrivial center of a finite p-group directly from the conjugation classes.