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

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!

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Are you assuming that G is abelian - you don't state this, but have tagged abelian groups - if not the dihedral group of order $2q$ is a counterexample. –  Mark Bennet Oct 22 '12 at 10:34
i didnt mean to state G as abelian, apologies!! –  d13 Oct 22 '12 at 10:41
@MarkBennet What is the normal order 2 subgroup of the dihedral group? –  Phira Oct 22 '12 at 10:59
@MarkBennet: How is the dihedral group of order $2q$ a counterexample? I don't see any normal subgroup of order $2$. –  Tara B Oct 22 '12 at 10:59
@Phira: Snap! I expect Mark misread the question, as I originally did, as being about a normal subgroup of index $2$. –  Tara B Oct 22 '12 at 11:00
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Choose a generator $g$ of $G/D$ and look at the order of $g$ and $ga$ in $G$.

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if ga=ag then |ga|=|g|*|a| (then im done), but how do i know if ga=ag since normal means gag^-1 is not equal a. –  d13 Oct 22 '12 at 10:37
Normal means that $gag^{-1}$ is in $D$. You can easily check that it has to be $a$ after all. Your first sentence is not entirely true, it is only true provided the orders are coprime. –  Phira Oct 22 '12 at 10:57
yes yes |ga|=|g||a| only if gcd(|a|,|g|)=1 i forgot to mention it! –  d13 Oct 22 '12 at 11:11
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.
I may be missing something, but as far as I can see, this only shows that every element of $G$ commutes with $y$, which isn't by itself enough to show that $G$ is abelian. –  Tara B Oct 22 '12 at 11:05
@math-123: Going directly from $gyg^{-1} = y$ for all $g\in G\setminus H$ to '$G$ is abelian'. It would have been more accurate for me to say there is a gap in the reasoning rather than that it is incorrect. –  Tara B Oct 25 '12 at 21:05