Timeline for About the elements of bounded order in finite groups
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| Nov 16, 2020 at 9:55 | comment | added | Jeremy Rickard | Related answer: mathoverflow.net/questions/337477/a-question-on-p-group-bounds/… | |
| Nov 16, 2020 at 9:17 | comment | added | Alireza Abdollahi | [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]. | |
| Nov 16, 2020 at 9:03 | history | edited | Alireza Abdollahi | CC BY-SA 4.0 |
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| Nov 16, 2020 at 3:45 | comment | added | verret | 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 $n=2$. | |
| Nov 16, 2020 at 0:04 | comment | added | verret | The reference for $n=2$ is "C.T.C. Wall, On groups consisting mostly of involutions, Proc. Cambridge Phil. Soc., 67 (1970) 251-262." I think 0.75 is indeed the maximum. | |
| Nov 15, 2020 at 23:59 | comment | added | verret | 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... | |
| Nov 15, 2020 at 21:05 | comment | added | R. van Dobben de Bruyn | 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? | |
| Nov 15, 2020 at 17:08 | comment | added | Geoff Robinson | 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 $|G|$. | |
| Nov 15, 2020 at 15:07 | history | edited | YCor | CC BY-SA 4.0 |
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| Nov 15, 2020 at 12:03 | history | edited | YCor |
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| Nov 15, 2020 at 12:03 | history | edited | MSMalekan | CC BY-SA 4.0 |
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| Nov 15, 2020 at 11:31 | history | asked | MSMalekan | CC BY-SA 4.0 |