Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 3 / 2D / Solution

For a finite group, the conjugacy classes partition the group, and a class is a singleton exactly when its element belongs to the centre of a group. Since the centre is trivial, the class equation becomeswhere the ${\mathcal{C}}_j$ are all the nonidentity conjugacy classes. If the prime number $p$ divided every $|{\mathcal{C}}_j|$, reduction modulo $p$ would give $0=1$, because $p\mid |G|$. Therefore at least one of these class sizes is not divisible by $p$. Primality now givesIts size is greater than one because the centre is trivial. This proves the prime-to-p conjugacy class lemma.

The conclusion requires $p$ to be prime. The printed question does not explicitly impose this hypothesis. If arbitrary composite divisors are allowed, take the symmetric group $S_3$ and $p=6$: its centre is trivial, its two nonidentity conjugacy classes have sizes $3$ and $2$, and neither is coprime to $6$. Thus the argument proves the intended prime case and also identifies why the unrestricted literal reading is false.