2

I need to prove the following:

Let $G$ be a abelian group such that $|G| = 2p$ and $p$ Is a odd prime number. Prove $G$ is a cyclic group.

So far I was able to show that there must be atleast one element $x$ such that $o(x) > 2$. If $o(x) = 2p$, then $G$ Is cyclic.

But, shat happens if $o(x) = p$?

Any hints will be appericiated :)

Shaun
  • 44,997
aaa bbb
  • 330
  • A group with order $2p$ with an odd prime $p$ is either cyclic or the Dieder-group. – Peter Jan 09 '22 at 17:11
  • By Cauchy there is $y$ of order 2. Then $xy$ has order $2p$. – markvs Jan 09 '22 at 17:12
  • @Peter Yes, But how can I show It is a cyclic group? – aaa bbb Jan 09 '22 at 17:13
  • 2
    What @Peter tried to say is that the Dihedral group is not commutative. – markvs Jan 09 '22 at 17:15
  • 2
    It is a bit hard to answer this question because I don't know what I am allowed to use. For example, this follows directly from the Fundamental Theorem of Finite Abelian Groups. It also follows directly from Sylow theorems for primes $2$ and $p$. However, I am not sure you would be happy with proofs going like that! –  Jan 09 '22 at 17:19

2 Answers2

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.

Sourav Ghosh
  • 12,997
0

If $o(x)=p$ take the quotient $G/\langle x\rangle$. Based on the order you can determine what is quotient group $H$. Then, by commutativity, you have $G \cong \langle x\rangle \times H$. There is an element in that Cartesian product group with order $2p$.

Shannon Starr
  • 633