5
$\begingroup$

Let $G$ be a finite group and $n$ be a positive integer. Define $$X_n(G):=\{x\in G: x^n=1\}.$$ Define $$J_n=\left\{\frac{|X_n(G)|}{|G|}: G\,\text{is a finite group}\right\}.$$ Question. Does there exist a positive integer $n$, such that $J_n=\mathbb Q\cap[0,1]$?

Here $\mathbb Q$ is the set of rational numbers? The answer to the simpler question below is also desirable to me:

Does there exist $n\in\mathbb N$, and a family of finite groups $(G_n)_n$ such that

  1. $|X_n(G_k)|<|G_k|$ for all $k$,
  2. $0<\prod_{k=1}^\infty\frac{|X_n(G_k)|}{|G_k|}$?
Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ It may be helpful to note that $|X_{n}(G)| = |X_{d}(G)|$, where $d = {\rm gcd}(n,|G|)$, and that (by a Theorem of Frobenius), $|X_{d}(G)|$ is an integer multiple of $d$ when $d$ divides $|G|$. $\endgroup$
    – Geoff Robinson
    Commented Nov 15, 2020 at 17:08
  • 1
    $\begingroup$ Note that the second question is equivalent to asking if there exists $n$ such that $J_n \setminus \{1\}$ has arbitrarily large elements. Indeed, if it does, you may inductively choose $x_k \in J_n \setminus\{1\}$ such that $x_1\cdots x_k > \frac{1}{2}$ for instance. The highest I have been able to produce is $\frac{3}{4} \in J_2$ exhibited by $D_4$; have you been able to get any higher? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 15, 2020 at 21:05
  • 1
    $\begingroup$ If you allow $n$ to vary, you can definitely get as close to $1$ as you wish. Just take for example $G=\mathrm{AGL}(1,q)$ and $n=q-1$. Then $|G|=q(q-1)$ and $|X_n(G)|=(q-1)^2$, so the ratio is $\frac{q-1}{q}$. With $n$ fixed, I am not sure. I remember seeing a paper dealing with the case $n=2$, and there it definitely does not get arbitrarily close to $1$. I'll see if I can find the reference... $\endgroup$
    – verret
    Commented Nov 15, 2020 at 23:59
  • 2
    $\begingroup$ Based on some computer experiments, I would guess that, for every $n$, there is a definite gap in $J_n$ just below $1$, just like in the case $n=2$. $\endgroup$
    – verret
    Commented Nov 16, 2020 at 3:45
  • 4
    $\begingroup$ [1] Thomas J. Laffey, The Number of Solutions of $x^3 = 1$ in a 3-Group, Math. Z. 149, 43 -45 (1976). [2] THOMAS J. LAFFEY, The number of solutions of $x^p = 1$ in a finite group, Math. Proc. Camb. Phil. Soc. (1976), 80, 229. $J_3 \subset [0,\frac{7}{9}] \cup \{1\}$ follows from [1] and [2]; $J_2 \subset [0,\frac{3}{4}] \cup \{1\}$; it is mentioned in [1] that ``it is easy to check that if $G$ is a 2-group not of exponent 2, then $X_2(G)\leq \frac{3}{4}|G|$"; the case $G$ is not a 2-group follows from [2]. $\endgroup$
    – Alireza Abdollahi
    Commented Nov 16, 2020 at 9:17

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .