The group has order $3\times 7$, and we know there exists at least one Sylow $3$-subgroup and at least one Sylow $7$-subgroup (here we can "just" use Cauchy's theorem that if $p\mid |G|$ there exists a subgroup of order $p$ in $G$. Sylow generalizes this to any power of $p$). Moreover, (one of) the Sylow $7$-subgroup has index $3$, and since $3$ is the least prime divisor of $21^1$, this subgroup must be normal, so it is unique; call it $Q$. Let $P$ be a Sylow $3$-subgroup. Then $QP$ is all of $G$: since $(3,7)=1$, $Q\cap P=1$, and hence $|QP|=|Q\cap P||QP|=|Q||P|=21$. It follows $G$ is a semidirect product $G=P\rtimes Q$, and we know that ${\rm Aut}\; Q=C_6$. Since $3$ divides $6$, we can produce nontrivial group homomoprhisms $\eta:C_3\to C_6$; and they will all give rise to a (unique) nonabelian group of order $21$.
$1$. If $G$ is a group of finite order, and $H$ has index $p$ where $p$ is the least prime divisor of $G$, then $H$ must be normal: the action of $G$ on the (say left) cosets of $H$ gives us a morphism $\eta:G\to S_p$ that induces an injection $\bar\eta:G/\ker\eta\to S_p$. This means the index $|G:\ker \eta|$ which equals $ |G:H||H:\ker \eta|=p|H:\ker \eta|$ divides $p!$, or that $|H:\ker \eta|$ divides $(p-1)!$. But $|H:\ker\eta|$ divides $G$, so if $|H:\ker \eta|\neq 1$, we would have prime factors from $G$ in $(p-1)!$, which has prime factors strictly smaller than $p$. It follows that $|H:\ker \eta|=1$, that is $H=\ker \eta\lhd G$, being a kernel of a morphism.
