= Solution
{c}
Choose a <normal subgroup> $N$ of prime order $p\in\{5,7\}$ as in <6/c/ii>, and choose a <subgroup> $H$ of the other prime order $q$. Their intersection is trivial by <Lagrange's theorem>. Both are <cyclic groups>, and <6/b/ii> gives $|\operatorname{Aut}(N)|=p-1$.
The <image of a group homomorphism> $\pi:H\to\operatorname{Aut}(N)$ from <6/b/iii> has order dividing $q$, because its <kernel of a group homomorphism> partitions $H$ into equally sized <cosets>. Its order also divides $p-1$ by <Lagrange's theorem>. But
$$
\gcd(7,4)=\gcd(5,6)=1,
$$
so $\pi$ is trivial in either case. By <6/b/iii>, $\langle N,H\rangle\cong C_p\times C_q$ has order $pq=35$ and therefore equals $G$.
If $n$ and $h$ generate $N$ and $H$, the element $(n,h)$ has order $\operatorname{lcm}(p,q)=pq=35$, since its $k$th power is the identity exactly when both $p$ and $q$ divide $k$. Thus this <direct product of groups> is cyclic, and
$$
\boxed{G\cong C_{35}.}
$$
Back to article page
