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Let $G$ be a finite group and $p\mid |G|$ be prime. Can $G$ have exactly $p-1$ elements of order $p$? (except trivial groups which are isomorphic to $\Bbb Z_p$)

I remember something similar to it. I don't remember exactly the conditions. it said either there's no elements of order $p$ or there's at least $2p-2$ elements. I'm trying to find the exact conditions.

Do you rememeber any theorem with a similar statement?

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If you only had tried the most obvious example... – Lord_Farin Apr 17 '13 at 17:29
Confused about the edit, Cauchy's theorem says whenever $p | |G|$, there is an element of order $p$. – muzzlator Apr 17 '13 at 17:36
up vote 1 down vote accepted

If the group is simple, then this is true.

That's because if there are only $p-1$ elements of order $p$ then they, along with the identity, form a normal subgroup of your group.

Any two cyclic subgroups of order $p$ cannot have non-trivial intersection, so if $G$ is simple, it either must be $\mathbb Z_p$, or it has $0$ or at least $2p-2$ elements of order $p$.

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yes i think this was what I tried to remember!. thanks. – user59671 Apr 17 '13 at 18:04

What I call property below is: the number of order $p$ elements is $p-1$.

Yes, this is possible. Take $\mathbb{Z}_p$ or, more generally, $\mathbb{Z}_{p^km}=\mathbb{Z}_{p^k}\oplus \mathbb{Z}_{m}$ for any $k\geq 1$ and any positive integer $m$ relatively prime with $p$. In other words, any cyclic group such that $p$ divides its order.

Note that a finite Abelian group $G$ has the desired property if and only if $p\mid |G|$ and if the $p$-Sylow of $G$ is cyclic.

For a non-Abelian example, take the dihedral group $D_{2p}$ with $p\geq 3$.

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any other example? – user59671 Apr 17 '13 at 17:36
@julien How about $\mathbb Z_{p^k}$? – Erick Wong Apr 17 '13 at 17:55
@ErickWong Oops. Thanks. – 1015 Apr 17 '13 at 19:44

When $p$ is a prime, the number of elements of order $p$ in any finite group $G$ s an integer multiple of $p-1.$ We may define an equivalence relation on the set $S$ of elements of order $p$ in $G$ via $x \sim y$ if and only if $\langle x \rangle = \langle y \rangle.$ Each equivalence class contains $p-1$ elements.

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$\Bbb Z_p$ has $p-1$ elements of order $p$.

Look at $\Bbb Z_p^2$ to see an example of where you don't have $p-1$ elements of order $p$ when $p | |G|$

Edit for your edited question:

For any group $G$, Cauchy's theorem says that if $p \ | \ |G|$, then there is an element of order $p$. An element $g$ of order $p$ generates a subgroup of $G$ given by $\langle g \rangle \cong \Bbb Z_p$. Each non-identity element in $\langle g \rangle$ will also have order $p$. This shows that every group of order divisible by $p$ has at least $p-1$ elements of order $p$.

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This actually has $p^2-1$ elements of order $p$. – Lord_Farin Apr 17 '13 at 17:30
Oh, misread the question. Thought it was asking "Must $G$ have $p-1$ elements" because otherwise $\Bbb Z_p$ would be a simple counterexample – muzzlator Apr 17 '13 at 17:31
@muzzlator You may edit your answer . – srijan Apr 17 '13 at 17:33

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