Show that a group $G$ of order 42 has a normal cyclic subgroup of order 21.
What I did so far is using Sylow's theorem to show that $G$ has a unique 7-sylow subgroup $S(7)$ (which is normal) and {1 or 7} of 3-sylow subgroup $S(3)$. If $n_3$ (the number of $S(3)$) is 1, Then $H:=S(7) \times S(3)$ is normal and also cyclic. However, the problem occurs when $n_3 = 7$. In this case, now $H$ can be written as the semi-direct product, namaly $S(3) \rtimes_\phi S(7)$. My question is how to show this $H = S(3) \rtimes_\phi S(7)$ is also cyclic? Thank you in advance.