 Solution

Let $P$ and $Q$ be subgroups of orders five and seven, which exist by i. Their normalizers contain $P$ and $Q$, respectively. By Lagrange's theorem, $N_G(P)$ has order $5$ or $35$, and $N_G(Q)$ has order $7$ or $35$. If either has order $35$, its subgroup is normal by iv.

Suppose neither is normal. Then iv gives at least seven distinct subgroups of order five and at least five distinct subgroups of order seven. Distinct subgroups of the same prime order intersect only in the identity, and a subgroup of order five and one of order seven also intersect only in the identity by Lagrange's theorem. Thus $G$ would have at leastnonidentity elements, contradicting $|G|-1=34$. Hence at least one of $P,Q$ is a normal subgroup.