Timeline for Proving some finitely presented groups being unequal
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 17, 2019 at 9:10 | vote | accept | Daniel | ||
| Jun 14, 2019 at 8:13 | answer | added | Daniel | timeline score: 2 | |
| Jun 14, 2019 at 7:17 | comment | added | Derek Holt | It might help to observe that the relation $(ab)^n=(ba)^n$ implies that $(ab)^n$ commutes with $a$ and $b$, and hence $(ab)^n$ is central. So, by working modulo $\langle (ab)^n \rangle$, it would be enough to prove that no nontrivial power of $ab$ is central in $D(k,l,n)$ (which is not true when $k=l=2$ and $n$ is even, but I suspect it is true otherwise). | |
| Jun 14, 2019 at 0:23 | comment | added | Andrea Marino | I would turn this into an answer tomorrow! It can actually be detailed. | |
| Jun 14, 2019 at 0:07 | comment | added | Andrea Marino | A more combinatorial approach. Take the free group on 3 generators $F_3$ and the canonical map to $G(l,k,n)$. Let $T$ be the kernel. It is generated by $a^l, b^k, \sigma:=(ab)^n(a^{-1}b^{-1})^n$.We want $\tau:=(ab)^m(a^{-1}b^{-1})^m \not \in T$. Call $G=\langle a^l, b^k \rangle , H= \langle\sigma \rangle $. Evidently $\tau \not \in G$, so that possibly $\tau = g_1 h_1 \ldots h_n g_n$; it may have different endings, but there is one $h$. I then claim that $\tau$, when simplified, contains $b(ab)^{n-1}(a^{-1}b^{-1})^{n-1}a$, morally because high and low exponents alternate and cannot simplify. | |
| Jun 13, 2019 at 23:33 | comment | added | Andrea Marino | As it seems a reasonable guess, I would try to find a representation of the n- group without the m condition. There are at least two ways to think about this: 1. Find a subgroup of S_n that makes the work. 2. Find a subgroup of matrices that makes the work. Without compitations, you can think about a group action (linear or not) where m-condition is not satisfied. | |
| Jun 13, 2019 at 14:58 | comment | added | YCor | OK, it's sometimes called "isomorphic as marked groups". | |
| Jun 13, 2019 at 14:52 | comment | added | Daniel | Well, by "unequal" I meant that sending generators on generators ($a\mapsto a'$, $b\mapsto b'$) does not extend to an isomorphism. Anyway, that's why I added the subsequent clarifying sentence. | |
| Jun 13, 2019 at 14:45 | comment | added | YCor | "Unequal" is not what you mean. The subgroups $\langle (12)\rangle$ and $\langle (23)\rangle$ of the symmetric group $S_3$ are isomorphic but not equal. | |
| Jun 13, 2019 at 14:45 | review | First posts | |||
| Jun 13, 2019 at 15:32 | |||||
| Jun 13, 2019 at 14:43 | history | asked | Daniel | CC BY-SA 4.0 |