Suppose we have a finite group $G$ and $d\in \mathbb N$ is a divisor of $|G|$. We define the set $E_d= \{g\in G : g^d =1\}$. Prove that $d$ is also a divisor of $|E_d|$.

So far I proved that $E_d=\displaystyle \bigcup_{g\in G : o(g)\vert d} \langle g \rangle$ but I didn't know how to continue from here and if this equality is helpful or not. I will appreciate any help. Thank you.


This is a theorem of Frobenius. He proved:

Theorem (Frobenius 1903): Let $d$ be a divisor of $|G|$ and $a_d$ be the number of elements of order dividing $d$. Then $a_d$ is divisible by $d$.

Note that $a_d=|E_d|$. A proof of this theorem is given, for example, by K. Brown here in Corollary $1.6$. The proof uses that $$ a_d=\sum \mu (H,K) |H|, $$ where $H$ and $K$ range over subgroups of order dividing $d$, obtained by Moebius inversion formula applied to the equations $$ |H|=\sum_{K\le H}f(K), $$ where $f(K)=0$, if $K$ is not cyclic, and $f(K)=\phi(k)$, if $K$ is cyclic of order $k$. For a detailed proof see the article On the Moebius function of a finite group by Hawkes et al, Theorem $6.3$.

Edit: The comments refer to an ealier answer.

  • $\begingroup$ $\sum a_kb_k$ needn't be a multiple of $\sum a_k$ $\endgroup$ – anon May 29 '15 at 11:41
  • $\begingroup$ What is an easy argument for the first statement? $\endgroup$ – Tobias Kildetoft May 29 '15 at 11:41
  • $\begingroup$ @TobiasKildetoft The map $x\mapsto\langle x\rangle$ from $\{$elements of order $d\}$ to $\{$cyclic subgroups of order $d\}$ is a $\phi(d)$-to-one map. $\endgroup$ – anon May 29 '15 at 11:42

Consider the set of $d$-tuples

\begin{eqnarray} S=\{(g_1, \cdots, g_d)|g_1g_2\cdots g_d=e, g_1, g_2, \cdots, g_d\text{ are not periodic with periodicity}>1\text{ and }<d\}\end{eqnarray}

Note that $|S|$ is divisible by $d$. Consider the $\mathbb{Z}_d$-action on $S$ by cyclically permuting the entries in a tuple, i.e. the generator $1\in\mathbb{Z}_d$ takes $(g_1, g_2, \cdots, g_d)$ to $(g_2, g_3, \cdots, g_d, g_1)$. There are two types of orbits, namely those tuples in $E_d$ and the orbits consisting of tuples with some entries distinct. Each tuple in $E_d$ forms an orbit itself, while each orbit of the other type consists of $d$ tuples. By class equation,

\begin{eqnarray}|S|=|E_d|+\text{#tuples with some entries distinct}\end{eqnarray} Note that the set of tuples with some entries distinct is a disjoint union of orbits each of which consists of $d$ tuples. Thus $|E_d|$ is divisible by $d$.

Remark: This is just an adaption of the proof of Cauchy's theorem.

  • 1
    $\begingroup$ "Note that the set of tuples with some entries distinct is a disjoint union of orbits each of which consists of d tuples" - Doesn't thsi assume that $d$ is prime? $\endgroup$ – Hagen von Eitzen May 29 '15 at 13:45
  • $\begingroup$ Let $d=4$, $a,b\in G$ such that $a\neq b$ and $(ab)^2=1$. Then involution in $\mathbb{Z}_d$ fixes tuple $s:=(a,b,a,b)$. Hence $\mathbb{Z}_d$-orbit of $s$ has less than $d$ elements. For example, we can take $G=S_4$, $a=(1,2,3),b=(2,3,4)$. Then $ab=(1,3)(2,4)$ and $a\neq b$, $(ab)^2=1$. $\endgroup$ – Alex W May 29 '15 at 15:42
  • $\begingroup$ @Hagenvoneitzen Alexw Thanks for pointing out the mistake on my argument. I did overlook the possibility of orbits with less than $d$ elements. After removing those problematic orbits (I.e. Those with entries exhibiting periodicity), the number of the remaining tuples is still divisible by $d$, and my argument still goes through. I'll supply more details later as now it's bedtime. $\endgroup$ – Alex Fok May 29 '15 at 16:21
  • $\begingroup$ Why $|S|$ is divisible by $d$ ? $\endgroup$ – Alex W May 29 '15 at 16:41
  • 1
    $\begingroup$ And yet, why $|S|$ is divisible by $d$? How to prove it in the beginning? Of course, number of non-fixed under $\mathbb{Z}_d$ tuples in $S$ is divisible by $d$. But why number of fixed points in $S$ relating $\mathbb{Z}_d$ is divisible by $d$? We did not know this until we prove theorem. Or why the number of "problematic" tuples is divisible by $d$? $\endgroup$ – Alex W Jun 2 '15 at 17:37

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