How many Sylow-$ 3$ subgroup does $G$ have? Let $G$ be a noncyclic group of order $21$.  How many sylow-$3$ subgroup does G have?
The possible orders of Sylow $3$ subgroups is $1, 7$. But how to check the exact number?

If it is $1$, $G$ must be cyclic as Sylow-$7$ subgroup is uniqe so it must be $7$.

Notice that $n_7$ must be equal to $1$, so it has a uniqe sylow-$7$ subgroup.
As you said $n_3\in \{1,7\}$, if $n_3=1$;
it has also normal subgroup of order $3$ which means $G=HK$ and $H\cap K=1$ and $H,K$ is normal in $G$ which means that $G$ is abelian.
Since order of $H,K$ is relativly prime, $G$ must be cyclic which is a contradiction.
