# Number of elements of order coprime to $p$ in a finite group

Let $$G$$ be a finite group. It is easy to prove that the elements of $$G$$ with odd order are in odd number. Indeed, if, for every divisor $$d$$ of the order of $$G$$, $$r_{d}$$ denotes the number of elements of $$G$$ with order $$d$$, then the number of elements of $$G$$ with odd order is $$1 + \sum_{d} r_{d}$$, where $$d$$ ranges over the odd divisors of $$\vert G \vert$$ such that $$d \geq 3$$; but $$r_{d}$$ is divisible by $$\varphi (d)$$ (Euler function) and, for a naturel number $$n \geq 3$$, $$\varphi (n)$$ is even, thus it is well true that the elements of $$G$$ with odd order are in odd number.

So, my question is : let $$G$$ be a finite group and $$p$$ a prime number; is it known if it is necessarily true that the elements of $$G$$ with order coprime to $$p$$ are in number coprime to $$p$$ ?

I found some lemmas and, if I am not wrong, these lemmas make it possible to prove that the answer is "yes" for every group of order $$\leq 431$$. Since my proof of this meager result is very long, I will only sketch it if I don't get a better answer.

Edit (August 8, 2021). I discover that this question is Exercise 28 b) in Bourbaki, Algèbre, ch. I (Paris, 1970), § 6, p. I.139.

• @HereToRelax Every cyclic subgroup of order $d$ has exactly $\varphi(d)$ generators. Apr 28, 2021 at 16:06
• Isn't it easier, to prove the result mentioned in the first paragraph, that every element of odd order greater than $1$ can be paired with its inverse (which is necessarily different from itself), yielding an even number; and then you also have the identity of order $1$, giving you an odd total? Apr 28, 2021 at 16:48

Let $$\Omega_p(G)$$ denote the set of elements of $$G$$ of order coprime to $$p$$, and pick a $$p$$-Sylow subgroup $$P$$. We may let $$P$$ act on $$\Omega_p(G)$$ by conjugation, in which case the orbits’ sizes are powers of $$p$$.

The only orbits of size not divisible by $$p$$ then are the fixed points, which are precisely the elements of

$$C_G(P)\cap\Omega_p(G)=\Omega_p(C_G(P)),$$

so we have a congruence

$$|\Omega_p(G)|\equiv |\Omega_p(C_G(P))| \pmod p$$

Either $$C_G(P)$$ is strictly smaller than $$G$$ or $$P$$ is central. Thus, we may apply induction, with our base cases those groups $$G$$ with central $$p$$-Sylow subgroup $$P\le Z(G)$$. Then we can prove

Lemma. If $$G$$ has a central $$p$$-Sylow subgroup $$P$$ then $$\Omega_p(G)$$ is a transversal for $$G/P$$.

Proof. First we show no two elements $$x,y\in\Omega_p(G)$$ represent the same coset. If $$xP=yP$$ then $$y=xu$$ for some $$u\in P$$. Taking orders of both sides yields $$|y|=|xu|=|x||u|$$ since $$x,u$$ commute and have coprime orders. Since $$p\nmid|y|$$ and $$|u|$$ is a power of $$p$$, this forces $$|u|=1$$, hence $$u=e$$, hence $$x=y$$ are the same representative.

cont’d. Second we show every $$g\in G$$ is in $$\Omega_p(G)P$$. By Chinese Remainder Theorem for cyclic groups we can write an internal direct product $$\langle g\rangle=\langle g^\alpha\rangle\times\langle g^\beta\rangle$$ with $$g^\alpha\in\Omega_p(G)$$ and $$g^\beta\in P$$. Then we should also be able to write $$g=(g^\alpha)^\delta(g^\beta)^\gamma\in\Omega_p(G)P$$.

The lemma implies, in our base cases where there is a central Sylow subgroup, that $$|\Omega_p(G)|=[G:P]$$ is coprime to $$p$$.

And these are my considerations.

Let $$G$$ be a finite group of order $$n$$ and let $$p$$ be a prime and $$p\mid n$$.

Denote by $$\Omega_p(G)=\{x\in G\,\mid\,{\rm gcd}(|x|,p)=p\}$$ and $$\Omega_{p'}(G)=\{x\in G\,\mid\,{\rm gcd}(|x|,p)=1\}$$. Since $$\Omega_{p}(G)\cup\Omega_{p'}(G)=G$$ and $$\Omega_{p}(G)\cap\Omega_{p'}(G)=\varnothing$$ it follows that both numbers $$|\Omega_{p}(G)|$$ and $$|\Omega_{p'}(G)|$$ are not divisible by $$p$$ or divisible by $$p$$ at the same time.

Theorem. If $$G$$ is a finite group of order $$n$$ and $$p$$ is a prime and $$p\mid n$$, then $$|\Omega_{p}(G)|$$ is not divisible by $$p$$.

Proof hints:

1. Let $$P$$ be a Sylow $$p$$-subgroup of $$G$$. If $$P\leq Z(G)$$, then $$\Omega_{p}(G)=(g_1P\setminus\{g_1\})\cup\ldots\cup(g_sP\setminus\{g_s\})$$ where $$s=|G:P|$$ and $${\rm gcd}(|g_i|,p)=1$$. Then $$|\Omega_{p}(G)|=|G|-|G:P|$$.

2. The group $$P$$ acts on the set $$\Omega_{p}(G)$$ by conjugation. Then $$\Omega_{p}(G)$$ is a union of orbits under the action of $$P$$ and each such orbit contains $$|P:C_P(x)|$$ elements where $$x$$ lie in the orbit.

3. Consider orbits consisting of a single element (for them $$|P:C_P(x)|=1$$). Let $$H=\{x\in G\,\mid\,|P:C_P(x)|=1\}=\{x\in G\,\mid\,[x,P]=1\}.$$ Let $$O(x_1),\ldots,O(x_t)$$ be the orbits of size $$|P:C_P(x_i)|>1$$. Then $$\Omega_{p}(G)=O(x_1)\cup\ldots\cup O(x_t)\cup\Omega_{p}(H)$$ and $$|\Omega_{p}(G)|=|O(x_1)|+\ldots|O(x_t)|+|\Omega_{p}(H)|.$$ Since $$Z(P)\leq H$$, $$|H|$$ is divisible by $$p$$.

4. If $$H\neq G$$, then by induction $$|\Omega_{p}(H)|$$ is not divisible by $$p$$.

5. If $$H=G$$, then $$P\leq Z(G)$$ (see case 1)