TITLE: Group of order 21 and subgroup of order 7
QUESTION [5 upvotes]: I have this problem and I don't know how to solve it, because in this part of the book I only have normality and the isomorphism theorems.
Let $G$ a group and $|G|=21$ assume that $a \in G$ and $|a|=7.$ Prove that $A = 
 \langle a \rangle$, the subgroup generated by $a$, is normal in $G$. i.e $$A \lhd G$$
At this moment of the book I CAN'T use Sylow theorems so I dont know how to do this without it.
Thanks.

REPLY [6 votes]: Suppose that $H$ and $K$ are distinct subgroups of $G$ with $|H| = |K| = 7$. Then, since $7$ is prime, we must have $H \cap K = 1$, and so
$$|HK| = \frac{|H||K|}{|H \cap K|} = \frac{49}{|H \cap K|} = 49$$
This obviously can't happen in a group containing only $21$ elements. Therefore $\langle a \rangle$ must be the only subgroup of order $7$.
This implies that $\langle a \rangle$ is normal (indeed, characteristic).
Proof: any automorphism $\phi : G \to G$ must map $\langle a \rangle$ to a subgroup of the same size, hence must map $\langle a \rangle$ to itself. Conjugation by $g$ is an automorphism, hence $g\langle a \rangle g^{-1} = \langle a \rangle$ for all $g \in G$.