Prove that G is cyclic if normal subgroup is of order 2

Question

Let $|G|=2q$ for $q\geq3$ and $q$ is prime, and it has a normal subgroup of order $2$. Prove that $G$ is cyclic.

Here is what I did:

Let $D$ be a normal subgroup of $G$ and $|D|=2$, $D=\{e,a\}$ for some $a$. I found a quotient group of $|G/D|=q$. This implies that $G/D$ is cyclic therefore abelian.

But how does this imply that $G$ is cyclic? Because I know that a quotient group of cyclic is cyclic, but I don't know the other way around. Please help!

Thanks a lot!

Answer 1

Choose a generator $g$ of $G/D$ and look at the order of $g$ and $ga$ in $G$.

Answer 2

Since $H=\langle 1,y\rangle$ of order $2$ is normal, let $g\in G\setminus H$ be any element, then $\mathrm{ord}(gyg^{-1})=2$. Hence it is $y$, so the group $G$ is abelian, hence it is cyclic.