Timeline for A group associated to a pair of integers $(k,p)$ where $p$ is a prime number
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2020 at 17:55 | comment | added | LSpice | @YvesStalder, under this condition, we could just take $G_{k, p} = \mathbb Q/e\mathbb Z \subseteq \mathbb Q \times_{\mathbb Z} U(p)$, in which the image $\langle k + p\mathbb Z\rangle$ of $\phi$ embeds as $\mathbb Z/e\mathbb Z$, where $e$ is the order of $k$ modulo $p$. Since any $\alpha$ is trivial on $e\mathbb Z$, the "lifting" $\beta$ in this case is just the obvious projection $\mathbb Q \to \mathbb Q/e\mathbb Z$. | |
| Sep 16, 2020 at 20:13 | comment | added | Ali Taghavi | @YvesStalder yes this would be nice. BTW according to your answer I realize that perhaps the problem relate to "extention of Charactets" | |
| Sep 16, 2020 at 12:20 | comment | added | Yves Stalder | @AliTaghavi, would you be glad with an extension $\psi_{k,p}: \mathbb{Q} \to G_{k,p}$ of $\phi$ where $G_{k,p}$ contains the image of $\phi$ but not necessaryly $U(p)$? | |
| Sep 16, 2020 at 12:12 | comment | added | Yves Stalder | The property defining universal groups is a bit strange, since it does not guarantee uniqueness (up to isomorphism). I think that if $\psi_{k,p}:\mathbb{Q} \to G_{k,p}$ satisfies the property, then for any group $G'$, the map $\psi'_{k,p}:\mathbb{Q} \to G_{k,p}\times G', \ q \mapsto (\psi_{k,p}(q),e_{G'})$ still satisfies the property. I think one should require $\psi_{k,p}$ to be surjective. | |
| Sep 15, 2020 at 23:07 | comment | added | LSpice | @AliTaghavi, re, everybody seems to have different notational conventions. I mean the pushout $(\mathbb Q \times U(p))/\langle(1, k^{-1})\rangle$. It may be that no-one else uses this convention … maybe $\amalg_{\mathbb Z}$ would look better than $\times_{\mathbb Z}$? | |
| Sep 15, 2020 at 22:37 | comment | added | Ali Taghavi | Thank you for your answer. But I am not sure I can see an obvious answer for the second part. May be a slight modification of comment by @LSpice can give a complete answer | |
| Sep 15, 2020 at 22:35 | comment | added | Ali Taghavi | @LSpice Are there some thing missing in your comment? Are you sure about $\mathbb{Q}\times_Z U(P)$. en.m.wikipedia.org/wiki/Pullback_(category_theory) | |
| Sep 15, 2020 at 15:22 | comment | added | Yves Stalder | @LSpice, I agree, thanks ! | |
| Sep 15, 2020 at 14:05 | comment | added | LSpice | $\DeclareMathOperator\ord{ord}$I think we can just unroll your group, and define $G_{k, p} = \mathbb Q \times_{\mathbb Z} U(p)$, where $\mathbb Z \to \mathbb Q$ is obvious and $\mathbb Z \to U(p)$ sends $1 \mapsto k$. Then, for any $\alpha$ as in the problem, define $\beta(r, x) = \alpha(r)x$. | |
| Sep 15, 2020 at 7:03 | history | answered | Yves Stalder | CC BY-SA 4.0 |