 Solution

By Lagrange's theorem, every nonidentity group element has order $5$, $7$, or $35$. If some element has order $35$, its seventh and fifth powers have orders $5$ and $7$, respectively, and we are done.

Otherwise every nonidentity element has order $5$ or $7$. Distinct subgroups of a given prime number order intersect only in the identity: a shared nonidentity element generates both. If there were no element of order $5$, the $34$ nonidentity elements would therefore partition into sets of six, the nonidentity elements of the cyclic subgroups of order seven. This would force $6\mid 34$, a contradiction. If there were no element of order $7$, the analogous partition into sets of four would force $4\mid 34$, again a contradiction. Thus elements of both orders exist.