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


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?

  • $\begingroup$ 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


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 $p$.

  • $\begingroup$ 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

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.)


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