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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.

Choose a positive common denominator $d$ so that $x=a/d$ and $y=b/d$ with nonzero integers $a,b$. Put $h=\gcd (a,b)>0$. Every element of the generated subgroup has the form $(ma+nb)/d$, so it lies in $(h/d)\mathbb{Z}$. Conversely, Bezout identity gives integers $u,v$ with $ua+vb=h$, so $h/d$ belongs to the generated subgroup. ConsequentlyThis map is a bijective additive group homomorphism. It is the two-generator instance of a finitely generated subgroup of the rational additive group being an infinite cyclic group.

For the real example, take $x=1$ and $y=\sqrt{2}$. The group homomorphism $(m,n)\mapsto m+n\sqrt{2}$ from ${\mathbb{Z}}^2$ onto the generated subgroup is injective because $\sqrt{2}$ is an irrational number. Thus this subgroup is isomorphic to ${\mathbb{Z}}^2$. More directly, if a single real number $t$ generated it, then $1=at$ and $\sqrt{2}=bt$ for integers $a\ne 0,b$, giving $\sqrt{2}=b/a$, a contradiction. The choice $1,\sqrt{2}$ generates a noncyclic additive subgroup.