Normal subgroup of a finite group of order 21

Question

How many normal subgroups does a non abelian group $G$ of order 21 have other than the identity subgroup $\{e\}$ and $G$?

I think we can solve the problem using syllow's $p$-subgroups

Answer 1

Let $n_p$ be the number of $p$-Sylow subgroup of $G$. Note that $n_7\mid 3$ and hence $n_7=1$ or $3$. But since $n_7\equiv 1 {\pmod 7}$, then $n_7=1$ and hence $G$ has a normal subgroup of order $7$. Is there any other proper normal subgroup? If there is then it should be of order 3 and the only $3$-Sylow subgroup of order 3. But then $G$ is isomorphic to $\mathbb{Z}_7\times \mathbb{Z}_3$ which is Abelian. So $G$ must have only one normal proper subgroup.

Answer 2

According to syllow's second theorem

let $p \rvert o(G)$, where $G$ is finite group and $p$ is any prime number.Then, a sylow $p$-subgroup $H$ of $G$ is normal, if $H$ is a unique sylow $p$-subgroup of $G$

According to syllow's third theorem

The number of sylow $p$ subgroup in $G$ for a given number $p$ is of the form $1+kp$, where $k$ is some non-negative integer and $1+kp \rvert o(G)$

Now $21=3 \times 7$

Hence,It has $3$-sylow and $7$-sylow subgroups

According to syllow's third theorem

No of $7$-syllow subgroup are $1+7k$ and $1+7k \rvert 21$

hence $1+7k \rvert 3$

$\therefore 1+7k=1$

$\therefore$ There exist unique uniuqe $7$-sylow subgroup of $G$ which is normal

So I think answer is 1