= 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> becomes
$$
|G|=1+\sum_j |\mathcal C_j|,
$$
where 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 gives
$$
\boxed{\gcd(|\mathcal C_j|,p)=1.}
$$
Its size is greater than one because the centre is trivial. This proves the <prime-to-p conjugacy class lemma>.
\b[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.
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