0
$\begingroup$

Suppose $p$ is an odd prime and $G$ is a non-abelian group of order $2p$. Prove that $G$ contains an element of order $p$.

Attempt:

By Lagrange’s theorem, for all $x \in G$ , $o(x) = 1, 2, p , ~or~ 2p$

Case 1: There will not be any element of order $2p$ otherwise $G$ become cyclic and thus abelian (but given, $G$ is a non-abelian group).

Case 2: If there exist $x\in G$ such that $o(x)=p$ then there is nothing to prove.

Case 3: $e$ is the only element of order 1. $p\neq 1$, as $p$ is odd prime.

Case 4: Let there is an element $g\in G$ such that$o(g)=2$.

If none of the cases 1, 3, 4 hold, then proof will be completed. How to show that this case 4 is not possible?

$\endgroup$
2
  • $\begingroup$ Are you allowed to use Cauchy's theorem ? $\endgroup$
    – Peter
    Dec 31, 2017 at 12:12
  • $\begingroup$ @Peter Please solve without using Cauchy's theorem. $\endgroup$ Dec 31, 2017 at 12:20

2 Answers 2

4
$\begingroup$

Assume that all elements have order $1$ or order $2$.

Then $xyyx=xex=xx=e=xyxy$ for every $x,y\in G$ and consequently $yx=xy$.

So then the group must be abelian.

We conclude that some element must exist having order $p$ or $2p$.

But as you noted already the second possibility leads to a cyclic, hence abelian group.

Final conclusion: some element exists having order $p$.

$\endgroup$
2
  • $\begingroup$ Please elaborate the line $xyyx=xx=e=xyxy$ $\endgroup$ Dec 31, 2017 at 12:53
  • 1
    $\begingroup$ Since $y$ has order $2$ (or $1$) $yy=e$. Same for $x$ and for $xy$ $\endgroup$
    – drhab
    Dec 31, 2017 at 12:54
3
$\begingroup$

What you need to show is that $g^2=e$ can't be true for all $g\in G$ if $G$ is non-abelian. This holds because if $g^2=e$ for all $g\in G$, then

$$e=(xy)^2=xyxy\implies x=x^2yxy=yxy\implies yx=y^2xy=xy$$

(Ah, this is essentially the same as drhab's (edited) answer, just presented somewhat differently.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.