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. $0<\prod_{k=1}^\infty\frac{|X_n(G_k)|}{|G_k|}$?
$\endgroup$
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 $|G|$. $\endgroup$ 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? $\endgroup$ 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
    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
    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]. $\endgroup$ Nov 16, 2020 at 9:17

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.