1

I have G where $|G|=2p$ ; p is prime. $\exists a\in Z\left(G\right);\:a^2=e$. I need to prove that G is abelian.

Now, let's translate it into math. To prove that G is abelian, is in other words ti prove that $Z\left(G\right)=G$, but $Z\left(G\right)\in G$, so if I prove that \left|Z\left(G\right)\right|=\left|G\right|, I've done.

Now, by lagrange's theorem, $\left|Z\left(G\right)\right|$ can be 2, p or 2p (which I want to prove).

It cant be p, because there is $a\in Z\left(G\right)$, which means $a\in G$ so $a^{2}=e$, therefore $2 : |Z(G)|$.

So I'm left with 2 and 2p. But as I see, theoretical the center can be only 2 elements: $\left\{e,a\right\}$, because all the data I have that there is there some $a$, but $a^{-1}=a$ because $a^{2}=e$, so I have for sure 2 elements.

I dont see how can i prove that there is at least one another abelian element in Z(G), because the existence of two fits perfect eather, as I can see.

Can somebody give me a hint ? Can this exercize be solved like that or Am I in the wrong direction from the first place ?

Can somebody give me a hint?

Ilya.K.
  • 1,298
  • Look for my answer here : http://math.stackexchange.com/questions/739237/let-g-be-a-group-of-order-2p-where-p-is-a-prime-greater-than-2-then/739240#739240 – Ivan Di Liberti Aug 20 '15 at 11:29
  • It isnt exactly the same problem, but the only useful information that I see, that we can say that there exists in G an element whos order is p (despite the fact I dont see what is the explanation from discussion on the link you sended for that, but supose it is true). That doesnt mean yet it is in the center, wich I want to show. And secondly, I dont want use isomorphism here any way it is usable. – Ilya.K. Aug 20 '15 at 11:35

1 Answers1

3

By hypothesis there is a central element of order two. This means that $Z(G)$ has order at least $2$. Then $[G:Z(G)]$ is either $p$ or $1$. It cannot be $p$ for it would then be cyclic, and this would force $Z(G)=G$. So it is $1$, and $Z(G)=G$.

Pedro
  • 125,149
  • 19
  • 236
  • 403
  • why should'nt it be cyclic ? and when you write $[G:Z(G)]$ what do you mean, because I know this symbol as the number of cosets of Z(G) in that example. It is strange for me that you use it without that context. – Ilya.K. Aug 20 '15 at 11:38
  • @Ilya.K. If $G/Z(G)$ is cyclic then it is the trivial group $1$; that is, $G$ is abelian. You can always try to prove this, or search it here. – Pedro Aug 20 '15 at 11:40
  • Hm...I didnt learn that, interesting if there is another way to prove it. Is there ? – Ilya.K. Aug 20 '15 at 11:44
  • 1
    @Ilya.K. You should know that $G/Z(G)$ is cyclic implies $G$ is abelian. Maybe you have not seen this "theorem". – Nicky Hekster Aug 20 '15 at 11:44