Timeline for Automorphism group of $(\Bbb Z\times\Bbb Z) \rtimes \Bbb Z$
Current License: CC BY-SA 3.0
20 events
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| Aug 18, 2016 at 0:02 | comment | added | YCor | The group of positive automorphisms can be written as $B\ltimes U$ where $U$ consists of those automorphisms fixing identically $Z^2$ and the quotient. They fix $Z^2$ and map $c$ to an element $cw$ where $w\in Z^2$. They are inner when $w$ belongs to the derived subgroup $D$, so the outer part is $Z^2/D$ (which has order 2 when $c$ has order 4). As I said, this applies when the derived subgroup has finite index in $Z^2$, which holds iff 1 is not an eigenvalue of $c$. The specific remaining case need a specific argument. | |
| Aug 17, 2016 at 23:59 | comment | added | YCor | (continued) Anyway again assuming $Z^2$ is characteristic, we have in Aut a subgroup of index at most 2 containing inner automorphisms, preserving the quotient $Z$ (let's call them positive). Such elements act on $Z^2$ commuting with $c$. Actually, we get a subgroup $B$ of the automorphism group, fixing $c$ and acting on $Z^2$ as the centralizer of $c$. Actually $B/\langle c\rangle$ lies in the outer automorphism group (in case $c$ has order 4 this quotient is trivial). (continued) | |
| Aug 17, 2016 at 23:55 | comment | added | YCor | (continued) In case $Z^2$ is characteristic (which holds when the derived subgroup has finite index therein, which in turns holds when $c$ has order 3, 4, or 6), the action modulo $Z^2$ is an action on $Z$ which has Aut of order 2. Inner automorphisms do not reverse $Z$ but if some element of $GL(2,Z)$ conjugates $c$ to its inverse, then we get an outer automorphism. (To be continued) | |
| Aug 17, 2016 at 23:53 | comment | added | YCor | By the way I voted to move to MathSE, not to close the question. Anyway, here are the clues to solve such a question without asking the community: wonder if the normal $Z^2$ is a characteristic subgroup. That it would be the derived subgroup is a bit optimistic, but if the derived subgroup has finite index therein, then $Z^2$ is the inverse image of the torsion of the abelianization and hence is characteristic. (to be continued) | |
| Aug 17, 2016 at 23:45 | comment | added | YCor | When $c$ was of order 4 $a\mapsto b\mapsto a^{-1}$, Out is a Klein group of order 4, generated by, say, $(a,b,c)\mapsto (a,b^{-1},c^{-1})$ and $(a,b,c)\mapsto (a,b,ca)$. | |
| Aug 17, 2016 at 23:44 | comment | added | user44191 | When you say "either of the two generators of $SL(2, \mathbb{Z})$", are you referring to a standard presentation? If so, which one - the ones I know being $\langle x^4 = 1, x^2 = y^3\rangle$ and $\langle x^4 = 1, x^2 = (xy)^3\rangle$? | |
| Aug 17, 2016 at 23:42 | comment | added | YCor | I guess it was closed because it's an exercise... when the question was written, $c$ was an element of order 4. Of course in general there's more discussion. And no, the key question is not whether it's "a subgroup of $SL(3,Z)$" (which means nothing, or maybe whether it's isomorphic to a subgroup therein, but that would be a question coming after the answer. | |
| Aug 17, 2016 at 22:51 | history | reopened |
Stefan Kohl♦ Igor Belegradek Francois Ziegler Igor Rivin Yemon Choi |
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| Aug 17, 2016 at 22:38 | history | edited | SKShukla | CC BY-SA 3.0 |
added 115 characters in body
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| Aug 17, 2016 at 22:16 | comment | added | SKShukla | It seems it was judged to be too simple. I have tried calculating it and I don't see an obvious answer. One key question was whether it would be a subrgroup of $SL(3,Z)$. | |
| Aug 17, 2016 at 21:21 | comment | added | Igor Rivin | @AllenHatcher Agreed, I voted to reopen. | |
| Aug 17, 2016 at 19:23 | review | Reopen votes | |||
| Aug 17, 2016 at 22:53 | |||||
| Aug 17, 2016 at 19:04 | history | edited | SKShukla | CC BY-SA 3.0 |
added 51 characters in body; edited tags
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| Aug 17, 2016 at 17:36 | comment | added | Allen Hatcher | I don't see why the question was closed -- and just as I was finishing writing an answer! Maybe it can be reopened. | |
| S Aug 17, 2016 at 17:05 | history | unlocked | CommunityBot | ||
| S Aug 17, 2016 at 17:05 | history | locked | CommunityBot | ||
| S Aug 17, 2016 at 17:05 | history | closed |
YCor Jeremy Rickard Myshkin user21574 Stefan Kohl♦ |
Not suitable for this site | |
| Aug 17, 2016 at 12:49 | review | Close votes | |||
| Aug 17, 2016 at 17:05 | |||||
| Aug 17, 2016 at 11:15 | comment | added | Geoff Robinson | Note that $c^{4}$ is central in that group. | |
| Aug 17, 2016 at 11:00 | history | asked | SKShukla | CC BY-SA 3.0 |