Timeline for Infinitely generated subgroup of a finitely one (weird)
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 4, 2014 at 10:18 | answer | added | Geoff Robinson | timeline score: 7 | |
| Jun 4, 2014 at 9:38 | answer | added | ivanpenev | timeline score: 2 | |
| Jun 4, 2014 at 8:20 | comment | added | bof | @user42912 Have you considered Martin Brandenburg's hint? Do you know what the values of $x$ are? If not, that should be the next item on your agenda. You need to find out what $H$ is before you try to prove anything more complicated about $H$. | |
| Jun 4, 2014 at 7:54 | comment | added | M. Vinay | @user42912 If you consider the example of $a^{−1}ba \in H$, it's not all that weird that H may not be finitely generated, as $a$ is one of the two generators of $G$ but $a \notin H$, and $b$ alone cannot generate $a^{−1}ba$. Thanks, bof, for the correction! | |
| Jun 4, 2014 at 7:52 | comment | added | James | One way to show that $H$ is not finitely generated is to show that it is isomorphic to a group well known not to be finitely generated. Are you familiar with the quasicyclic groups? (Alternatively, you can show directly that any finite set in $H$ must be contained in a proper (in this case, cyclic) subgroup.) | |
| Jun 4, 2014 at 7:43 | comment | added | ivanpenev | @M.Vinay. Thank you, and my apologies. | |
| Jun 4, 2014 at 7:39 | comment | added | M. Vinay | @ivanpenev I doubt $H$ is generated by $b$. For $a^{-1}ba = \left(\begin{matrix}1 & 1/2\\0 & 1\end{matrix}\right) \in H$ (as its diagonal entries are $1$), but it is not generated by $b$ ($b^{-1} = \left(\begin{matrix}1 & -1\\0 & 1\end{matrix}\right)$). | |
| Jun 4, 2014 at 7:37 | comment | added | user42912 | @bof yes, you're right, I tried this strategy without success, I need some hint to begin to solve the question. | |
| Jun 4, 2014 at 7:23 | comment | added | user42912 | @ivanpenev yes, that's strange, because the question says $H$ is not finitely generated. | |
| Jun 4, 2014 at 7:21 | comment | added | user42912 | @bof I'm sorry, is it ok now? | |
| Jun 4, 2014 at 7:20 | history | edited | user42912 | CC BY-SA 3.0 | added 131 characters in body |
| Jun 4, 2014 at 7:12 | comment | added | Martin Brandenburg | More importantly, $a$ and $b$ don't lie in $H$, so that they don't generate $H$. By the way, the description of $H$ is incomplete: What is $x$? Not every real number ... | |
| Jun 4, 2014 at 7:10 | history | edited | egreg | CC BY-SA 3.0 | Image to text |
| Jun 4, 2014 at 7:09 | comment | added | David | Re: "really weird", perhaps your idea of generators is not quite accurate. Note that to say $H$ is generated by $a$ and $b$ means, roughly, that calculating all possible combinations of $a$ and $b$ will give you $H$ exactly, that is, $H$ and nothing more. But in your question, $a$ and $b$ generate $G$ which is more than $H$, not $H$ exactly. Not sure whether or not this is the issue which is confusing you but hope it helps. | |
| Jun 4, 2014 at 7:04 | history | asked | user42912 | CC BY-SA 3.0 |