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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!

Jess
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    Hint: Show that a normal subgroup of order $2$ is central. – Tobias Kildetoft Jan 21 '17 at 20:20
  • Any cyclic group is abelian, so your argument is longer than necessary. – David Wheeler Jan 21 '17 at 20:35
  • @TobiasKildetoft Would you show that by using the fact that H has 2 elements, $e$ and $h$, and since H is normal gH = Hg for all g in G, the set {gh,g} must equal {hg,g}, so gh=hg, and hence H is central? From that, how does that show G is abelian? – Jess Jan 21 '17 at 20:49
  • Yes, that is one way to show the hint. To follow it up: Are you familiar with the statement that if $G/Z(G)$ is cyclic then $G$ is abelian? – Tobias Kildetoft Jan 21 '17 at 20:53
  • @TobiasKildetoft That helps a lot, thank you! – Jess Jan 21 '17 at 20:55

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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.

Asinomás
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