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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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    @Phira I$n$deed. My folly.2012-10-22

2 Answers 2

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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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    yes yes |ga|=|g||a| only if gcd(|a|,|g|)=1 i forgot to mention it!2012-10-22
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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.

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    @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.2012-10-25