= Solution
{c}
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 $6\mid34$, a contradiction. If there were no element of order $7$, the analogous partition into sets of four would force $4\mid34$, again a contradiction. Thus \b[elements of both orders exist].
Back to article page
