Timeline for When is the semidirect product of an elementary abelian group and a cyclic group generated by two elements?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 16, 2018 at 12:59 | comment | added | Derek Holt | @YCor Yes of course! I got $p$ and $q$ the wrong way round. And the other direct factor should be. $C_p$ not $C_q$. | |
| Aug 16, 2018 at 10:09 | comment | added | YCor | @DerekHolt I think the more standard convention is to write $C_p\wr C_q$ ($C_q$ being the acting group)? | |
| Aug 16, 2018 at 7:02 | comment | added | Derek Holt | It may be of some interest to observe that the group is 2-generated if and only if it is a quotient of $C_q \times (C_q \wr C_p)$. | |
| Aug 16, 2018 at 5:42 | comment | added | Luc Guyot | @YCor Many thanks, I acknowledge my long-standing obliviousness. | |
| Aug 16, 2018 at 5:37 | vote | accept | Pat Devlin | ||
| Aug 16, 2018 at 5:13 | comment | added | YCor | @LucGuyot Derek already answered: according to whether $n_1\le 1$ or $n_1=2$, write $G=V\rtimes Z/qZ$ or $G=V\rtimes Z/pqZ$ with $V$ cyclic (i.e. with all multiplicities $\le 1$). Then use one generator of the right-hand cyclic group and one of the cyclic module. | |
| Aug 16, 2018 at 5:02 | comment | added | Luc Guyot | @YCor I understand why irreducible (modular) representations of $Z/qZ$ over $Z/pZ$ exist and why the condition on their multiplicities is necessary. But I fail to see why it is sufficient when $n_1 > 0$. | |
| Aug 15, 2018 at 17:29 | answer | added | Luc Guyot | timeline score: 7 | |
| Aug 15, 2018 at 16:56 | comment | added | Derek Holt | @PatDevlin You have some choice in the orders of the two generators. For example, if there are no trivial modules ($n_1=0$) then there is a choice between two generators of order $p$, or one of order $p$ and the other of order $q$. If $n_1=2$, then you must choose one generator of order $pq$ and the other can be of order $pq$ or $q$. There is even more choice when $n_1=1$. | |
| Aug 15, 2018 at 15:31 | comment | added | Pat Devlin | Thanks so much! I’ll have to bone up on representation theory to unpack this. As a quick possibly silly question, in your examples are your groups generated by elements of different orders? | |
| Aug 15, 2018 at 7:19 | comment | added | YCor | Write $V=(Z/pZ)^n$ as direct sum of irreducibles $V_i^{n_i}$ over $Z/qZ$, where the $V_i$ are pairwise non-isomorphic as $Z/qZ$-modules. Write $V_1=Z/pZ$ viewed as trivial $Z/qZ$-module. Write $m_i=n_i$ for $i>1$ and $m_i=n_i-1$. Then your group $G$ is generated by two elements if and only if $m_i\le 1$ for all $i$ (that is, $n_1\le 2$ and $n_i\le 1$ for $i>1$). Here $p$ is any prime and $q$ can be any power of a prime $\neq p$. I'll write details later but it's a good exercise. | |
| Aug 15, 2018 at 7:16 | comment | added | Tobias Kildetoft | One example where this happens is if the corresponding representation of $\mathbb{Z}/q\mathbb{Z}$ is irreducible, which can happen precisely when $n$ is the multiplicative order of $p$ mod $q$. | |
| Aug 15, 2018 at 6:22 | history | asked | Pat Devlin | CC BY-SA 4.0 |