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.
Best Answer
To sum up the comments:
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.