# Number of normal subgroups of a non abelian group of order $21$… CSIR December $2013$

Question is to :

Find the number of Normal subgroups of a nonabelian group $G$ of order $21$ other than $\{e\}$ and $G$.

What I have done so far is :

As $|G|=21=3\cdot7$ we have :

No. of sylow $3$ subgroups $1+3k$ dividing $7$ leaving out possibilities $1$ or $7$

No. of sylow $7$ subgroups $1+7k$ dividing $3$ leaving out only possibilities $1$.

So, we have a unique sylow $7$ subgroup and so it is normal.

I remember somehow that any normal group should come from a normal sylow subgroup or something like that.

S0, I prefer to conclude there is only one Normal subgroup for a non abelian group of order $21$.

Please let me know if this is true and please help me to fill that gap :

I remember somehow that any normal group should come from a normal sylow subgroup or something like that.

Thank you.

• I hope this post is usefull for you math.stackexchange.com/questions/117500/… – chuyenvien94 Dec 22 '13 at 11:42
• The answer is correct, since if the $3$ -Sylow subgroup is normal, you can show that $G$ is cyclic (and therefore abelian). – Ludolila Dec 22 '13 at 11:42
• @chuyenvien94 : Thank you for the link :) – user87543 Dec 22 '13 at 11:44
• The order of a non-trivial subgroup is either $3$ or $7$. Any group of order $3$ (in this case) is a $3$-Sylow subgroup... (Same goes for order $7$). So what other options you have? – Ludolila Dec 22 '13 at 11:48
• @srijan : yes yes.. No hard feelings... :) – user87543 Dec 22 '13 at 11:54

There is a normal subgroup of order $7$ (the $7$-Sylow subgroup). The $3$-Sylow subgroup is not normal, since otherwise $G$ would be cyclic (and therefore abelian).
Any non trivial (normal) subgroup is of order $3$ or $7$, and thus has to be a Sylow subgroup (in this particular case).
To conclude: $G$ has only one non-trivial normal subgroup.
• Well there is a connection between Sylow subgroups of a group and its normal subgroups, but it doesn't apply here (I think...). The connection is as follows: let $G$ be a group and $P$ is one of its Sylow-subgroups. Then if $N$ is a normal subgroup of $G$, the intersection $N \cap P$ is a Sylow-subgroup of $N$. – Ludolila Dec 23 '13 at 14:54