Let $G$ be a group and $H$ a normal subgroup of order 2. Prove that if $G/H$ is cyclic, then $G$ is abelian. Let $G$ be a group and $H$ a normal subgroup of order 2. Prove that if $G/H$ is cyclic, then $G$ is abelian.
I understand that if $G/H$ is cyclic, it is generated by a single element, i.e. $G/H = <xH>$, $x\in H$, and hence any $g \in G/H$ is
$g = (xH)^n, n\in \mathbb{Z}_{\geq 0}$, so $g = x^nH$. Thus for any 2 elements, $f,g \in G/H, fg = x^nHx^mH = x^{n+m}H = x^{m+n}H = x^mHx^nH = gf$, so $G/H$ is abelian.
Is this correct, and if so, can I use it to show G is abelian? What is the significance of H being order 2?
Any help appreciated!

A subgroup of order $2$: $\{e,h\}$ is normal if and only if $ghg^{-1}=h$ for all $g\in G$. In other words if $h$ commutes with every element of $G$.
Now use that $G/H$ is cyclic, in other words there is a $c$ such that $Hc^k$ covers all of $G/H$, in other words every element is of the form $c^k$ or $hc^k$.
Clearly any two such elements of this form commute.
