# $p$ is an odd prime, then prove that there is no group with exactly $p$ elements of order $p$.

Question

If $p$ is an odd prime, then prove that there is no group with exactly $p$ elements of order $p$.

Attempt

Assume that such a group $G$ exists.

If $x_1 \in G$ and if $|x_1|=p$ then $|{x_1}^{-1}|=p$

hence such elements occur in pairs,

$p$ being an odd prime implies $x=x^{-1}$ for some element x whose order is $p$. Otherwise there would be even number of elements of order $p$

$x=x^{-1}$

$\implies x^2=e$

$\implies |x|=2$

$\implies 2$ is an odd prime.

• The proof looks good to me. Aug 7, 2018 at 11:39
– Did
Aug 7, 2018 at 11:41
• Your proof is correct. Aug 7, 2018 at 11:59
• If $x^2=e$ then can't $x=e$ and hence $|x|=1$? The overall proof structure is still correct, but that last conclusion is a little bit overgeneral :) Aug 7, 2018 at 12:29
• @postmortes In the hypothesis, $|x|=p>1$. Aug 7, 2018 at 14:07

Other approach, pick an element of given ones of order $p$. But then all non trivial powers of this element give rise to elements of order $p$, hence there are at least $p-1$ of them. If $G$ has precisely $p$ elements of order $p$, then there must be precisely one other, giving also rise to $p-1$ elements of order $p$. Hence $2(p-1)=p$, so $p=2$ a contradiction. By the way: there are no groups (finite or infinite) with exactly $2$ elements of order $2$.
• How cauchy gurantees the existence of element of order $p$, When the Order of the group is not given how can you use cauchy's theorem. It is also possible that the Group is Infinite order. Aug 7, 2018 at 16:04
• Yes my approach is certainly for finite groups because of Cauchy. Should not have mentioned that. It is not clear from your post that it concerns also infinite groups. But my argument also applies to infinite groups. We do not need mr Cauchy since the elements of order $p$ are a given. Aug 7, 2018 at 16:11
• Your also did not mention that the group might be infinite. So why all this fuzz? If the group is finite then the existence of elements of order $p$ implies that the order of $G$ is divisible by $p$. Aug 7, 2018 at 16:17