Past exam of the options II course of the University of Oxford / 2012 / B (Group Theory)

The identity element satisfies $e\ast x=ex{e}^{-1}=x$. For $g,h\in G$,Both action axioms hold, so conjugation is a group action on $G$.

By definition, $g$ stabilizes $x$ exactly when $gx{g}^{-1}=x$. Multiplying on the right by $g$ gives the equivalent condition $gx=xg$. HenceThus the stabilizer in this action is the centralizer of the element.

Let $H=Z(x)$. The map $G/H\to C(x)$ sending $gH$ to $gx{g}^{-1}$ is well defined: multiplying $g$ on the right by an element commuting with $x$ leaves its conjugate unchanged. It is onto by the orbit definition. If $gx{g}^{-1}=hx{h}^{-1}$, then $h^{-1}g$ commutes with $x$, so $gH=hH$, proving injectivity.

Every coset has $|H|$ elements, so $|C(x)|=|G|/|H|$. ThereforeThis is the orbit-stabilizer theorem specialized to conjugation.

An element $x$ is central exactly when $gx=xg$ for every $g\in G$. By part (b), this is exactly the assertion that every $g$ stabilizes $x$. That means every conjugate of $x$ is $x$; conversely, a singleton conjugacy class has this property for every $g$.In particular, the singleton classes are in one-to-one correspondence with the elements of the centre.

Conjugacy is an equivalence relation: the identity element gives reflexivity, conjugation by $g^{-1}$ reverses conjugation by $g$, and composing conjugations gives transitivity. Its equivalence classes therefore partition the finite set $G$. Counting this disjoint union givesSeparating singleton classes from the others gives the class equation $|G|=|Z(G)|+{\sum }_{|C_i|>1}|C_i|$.

Write $|G|=p^n$, where $n\ge 1$. By part (c), each conjugacy-class size divides $p^n$ and hence is a power of $p$. Every nonsingleton class consequently has size divisible by $p$. The class equation impliesThe identity element lies in the centre, so the centre is nonempty. Its size is a positive multiple of $p$, and thereforeThis proves that a nontrivial finite p-group has a nontrivial centre.