Timeline for curious relation between orders of generators of a finite group
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jun 21, 2017 at 16:36 | vote | accept | stupid_question_bot | ||
| Jun 21, 2017 at 7:23 | answer | added | Fedor Petrov | timeline score: 5 | |
| Jun 21, 2017 at 6:58 | comment | added | user21230 | For $A_5$ group, $n=60$, but there is no element of order $30$ there. The orders of elements are $2,3,4,5$. The standard generators are of order $2,3$, for example $(1,2)(3,4)$, $(1,3,5)$ | |
| Jun 21, 2017 at 6:37 | review | Close votes | |||
| Jun 21, 2017 at 12:08 | |||||
| Jun 21, 2017 at 6:12 | comment | added | verret | Write $t=hgh^{-1}$. Then $G$ is generated by $t$ and $g$, which both have order $r$, and $gt^{-1}=ghg^{-1}h^{-1}$ has order $e$. In particular, $G$ is a quotient of $\langle g,t \mid g^r, t^r, (gt^{-1})^e\rangle$. When $r=2$, this is a dihedral group of order $2r$. When $r=3$ and $e=2$, this is $A_4$. | |
| Jun 21, 2017 at 6:01 | comment | added | Zach Teitler | There was a typo in my comment. Sorry. I meant: If $e,r \geq 3$ and $n>0$, then $e \geq \frac{n}{2+n-\frac{2n}{r}}$. And it also holds automatically if $e=2$ and $r \geq 4$. I guess the $r=2$ case, and maybe the $e=2,r=3$ case, could still be interesting! | |
| Jun 21, 2017 at 6:00 | comment | added | verret | When $r=2$, $g$ is an involution and so is $g^h$, so $G$ is generated by two involutions so must be a dihedral group. | |
| Jun 21, 2017 at 5:35 | comment | added | stupid_question_bot | @ZachTeitler Oh crap you're right. I guess I should have noticed that, hah! I guess at least the $r = 2$ case is still mildly interesting. At least this sort of validates my computations. | |
| Jun 21, 2017 at 5:16 | comment | added | Zach Teitler | For any real numbers, if $e,r \geq 3$ and $n > 0$, then $e \geq \frac{n}{2+n-\frac{2}{r}}$. | |
| Jun 21, 2017 at 5:11 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Jun 21, 2017 at 5:10 | comment | added | stupid_question_bot | @ycor yes sorry will clarify | |
| Jun 21, 2017 at 5:03 | comment | added | YCor | I guess $|g|$ is supposed to mean the order of $g$? | |
| Jun 21, 2017 at 4:54 | history | edited | stupid_question_bot |
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| Jun 21, 2017 at 4:19 | comment | added | Arturo Magidin | If $G$ is cyclic, then it is generated by $g$ of order $2$, and so $n=r=2$ and $e=1$, so the equality would also hold when $G$ is cyclic if $r=2$. | |
| Jun 21, 2017 at 4:14 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Jun 21, 2017 at 4:11 | comment | added | Arturo Magidin | Seems like a strong condition, and ex nihilo I wouldn't expect it to be true; but the condition you have on the generators may be stronger than it seems at first glance... | |
| Jun 21, 2017 at 4:09 | comment | added | stupid_question_bot | @ArturoMagidin Yes I suppose that is what I claim. Is that obviously wrong? The first example I checked was the case $G = D_{2k}$, where $g$ is a reflection, and $h$ is either a primitive rotation or another reflection which generates with $g$. In either case $g,hgh^{-1}$ generate $G$ iff $k$ is odd, and in that case $e$ is indeed equal to $k$, which is also the order of the commutator subgroup. | |
| Jun 21, 2017 at 3:55 | comment | added | Arturo Magidin | Wait... if $e$ is the order of $[g,h]$, then you necessarily have $e\leq n/2$: the order cannot be equal to $n$, since then $G$ would be cyclic and $ghg^{-1}h^{-1}$ would be trivial; and so must be a proper divisor of $n$, hence at most equal to $n/2$. Are you actually asserting that when $g$ is of order $2$, then the order $e$ of $[g,h]$ must be equal to $n/2$, and so that the commutator subgroup must be either of index $2$ or equal to all of $G$? | |
| Jun 21, 2017 at 3:11 | history | asked | stupid_question_bot | CC BY-SA 3.0 |