Let $G$ be a abelian group such that $|G| = 2p$ and $p$ Is a odd prime number. Prove $G$ is a cyclic group. [duplicate]
Solution 1:
$|G|=2p $ $(p \text { is a odd prime } ) $
Then by Cauchy's theorem of finite abelian group , $G$ has an element $a$ of order $2$ and $b$ of order $p$ .
As $a, b$ commutes and $gcd(2,p)=1 , |ab|=2p$
Hence, $G=\langle ab\rangle$
Alternative: $|G|=2p$
$H:2-SSG \cong \Bbb{Z_2}$ and $K:p-SSG\cong \Bbb{Z_p}$
And $H\cap K=\{e\}$ as $gcd(|H|,|K|)=1$
$H, K$ both are normal in $G$ , as $G$ is abelian.
Hence, $G=\Bbb{Z_2}\times \Bbb{Z_p}$ $(p\neq 2) $
And hence $G$ is cyclic.