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?

Already you have noticed number of Sylow-3 subgroup=$n_3 \in \{1,7\}$
and note that Sylow-$7$ subgroup is unique,hence normal.
If now Sylow-$3$ subgroup is unique then say $H$ and $K$ are the Sylow-$3$ and Sylow-$7$ subgroups respectively.
Then clearly $H\cap K=\{e\}$ and thus $|HK|=|G|=21\Rightarrow G=HK$
but we see that $\forall h\in H,k\in K, hkh^{-1}k^{-1}\in H\cap K=\{e\}$ because $h(kh^{-1}k^{-1})\in H$ as $H$ is normal and similarly $(hkh^{-1})k^{-1}\in K$ as $K$ is normal.So $hkh^{-1}k^{-1}=e\Rightarrow hk=kh$. More over both $H$ and $K$ are cyclic and hence abelian.
Thus $G$ is abelian.
Now we see that as $G$ must have element of order $3$ and element of order $7$ (By Cauchy's Theorem) let $o(a)=3,o(b)=7$ Since G is abelian  $ab=ba$ and also we see gcd $(3,7)=1$
Thus $o(ab)=3 \times 7=21=|G|\Rightarrow G$ cyclic which is a contradiction. Hence $n_3=7$
