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Suppose that every non-identity element of G has order p (prime). Then $\langle x\rangle = \{x^k:1\leqslant k\leqslant p\}$, and similarly for $y$. Suppose that $\langle x\rangle$ and $\langle y\rangle$ have a non-identity element in common, say $y^r=x^s.$ Let $t$ be the inverse of $r$ in $\mathbb{Z}_p^\ast$. Then $x^{st} = y^{rt} = y$, so that $y\in \langle x\rangle.$ Contrapositively, if $y\notin \langle x\rangle$ then $\langle x\rangle$ and $\langle y\rangle$ have no non-identity element in common.Yes it must work if p and q are co-prime because of the result that the order of an element divides the order of a group. If it doesn't work in the case p=q=prime, then I am confused about something in my textbook. We have a group G of order prime squared, non cyclic and it can be proved that the centre is non-trivial. Then we may choose x in Z(G) not the identity and y not in <x>. The book then states that <x> and <y> have only the identity in common. But clearly every non-identity element has order p.