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Let $k\in \mathbb{N}$ be a natural number and $p$ be a prime number with $p\nmid k$. (We thank Prof. Bartel for his comment on the latter non divisibility condition.) We denote by $U(p)=\{1,2,\ldots,p-1\}$ the mod-$p$ multiplicative group. Then we have a natural group homomorphism: $\phi:\mathbb{Z}\to U(p)$ which is the unique extension of semi group homomorphism $n\mapsto k^n \pmod p$ defined on semigroup of non negative integers.

Question: Is there a group $G$ containing $U(p)$ such that we have an extension $\psi:\mathbb{Q} \to G$ of $\phi$?

If the answer is yes, is there a universal group $G(k,p)$ with the following property:

There is an extension $\psi_{k,p}:\mathbb{Q} \to G_{k,p}$ of $\phi$. Moreover for any other extension $\alpha :\mathbb{Q} \to H\supset U(p)$ of $\phi$, there is a group homomorphism $\beta: G_{k,p} \to H$ with $\alpha =\beta \circ \psi_{k,p}$?

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  • $\begingroup$ @AlexB. Yes thank you. I revise it $\endgroup$
    – Ali Taghavi
    Commented Sep 14, 2020 at 19:45

1 Answer 1

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The group $U(p)$ is the group of units of the finite field $\mathbb{F}_p$, which is known to be cyclic. One can therefore identify $U(p)$ with the subgroup of $(p-1)$-th roots of unity in the group $S^1 = \{z\in \mathbb{C}: \, |z|= 1\}$. Now $k$ has the form $e^{i\frac{2\pi}{p-1}\ell}$ for some integer $\ell\in\{1,\ldots,p-1\}$, and it suffices to set $$ \psi : \mathbb{Q}\to S^1 \, ; \quad \frac ab \mapsto e^{i\frac ab\frac{2\pi}{p-1}\ell} \, . $$

I don't have the answer for the universal group $G_{k,p}$ until now.

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    $\begingroup$ $\DeclareMathOperator\ord{ord}$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 $\beta(r, x) = \alpha(r)x$. $\endgroup$
    – LSpice
    Commented Sep 15, 2020 at 14:05
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    $\begingroup$ @LSpice, I agree, thanks ! $\endgroup$
    – Yves Stalder
    Commented Sep 15, 2020 at 15:22
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    $\begingroup$ @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 $\times_{\mathbb Z}$? $\endgroup$
    – LSpice
    Commented Sep 15, 2020 at 23:07
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    $\begingroup$ 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. $\endgroup$
    – Yves Stalder
    Commented Sep 16, 2020 at 12:12
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    $\begingroup$ @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 $U(p)$? $\endgroup$
    – Yves Stalder
    Commented Sep 16, 2020 at 12:20

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