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Let $S$ be a monoid. A non-empty set $A$ is called a right $S$-act if there exists a map $A\times S\to A$ such that $(as)t=a(st)$ and $a1=a$ for all $s,t\in S$ and $a\in A$.

A sub-act of a right $S$-act $A$ is a non-empty subset of $A$ which is closed under the action of $S$.

A right S-act is called simple if it has no trivial subact.

For a prime $p$, $\mathbb{Z}_p$ is a right $\mathbb{N}$-act when we define the map $\mathbb{Z}_p \times \mathbb{N}\to \mathbb{Z}_p$ by $(\bar{x},n)\mapsto n\bar{x}$. My question is that: is $\mathbb{Z}_p$ a simple $\mathbb{N}$-act?

What I've tried: Let $A\neq \{ 0\}$ be a sub-act of $\mathbb{Z}_p$. Then there exists $\bar{0}\neq \bar{x}\in A$. Now if there exists $n_0\in \mathbb{N}$ such that $n_0 x\overset{p}{‎\equiv‎}1$, then $\bar{1}=n_0 \bar{x}\in A$. Then for all $n\in \mathbb{N}$, $\bar{n}=n\bar{1}\in A$ hence $A=\mathbb{Z}_p$.

Is there $n_0\in \mathbb{N}$ such that $n_0 x\overset{p}{‎\equiv‎}1$?

M.Ramana
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  • What is a simple act? – Kolja Oct 19 '22 at 13:48
  • $a1 = a$ doesn't make sense, as $a\in A$ and the image of an act should be an element of $S$, not of $A$. – Kolja Oct 19 '22 at 13:49
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    @Kolja A right $S$-act is called simple if it has no trivial subact. – M.Ramana Oct 19 '22 at 13:50
  • @Kolja The element $a1$ belongs to $A$ by the map and it is necessary to be $a$ by definition. – M.Ramana Oct 19 '22 at 13:53
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    Please change the codomain of the action map from $S$ to $A$ for $a1=a$ to make sense. – Geoffrey Trang Oct 19 '22 at 14:00
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    @GeoffreyTrang Sorry, I made a mistake. I've just edited. Thanks. – M.Ramana Oct 19 '22 at 14:03
  • I thought too much of this exercise. Am I right to assume that $\mathbb{Z}_p$ is the ring of integers modulo $p$, and not the ring of $p$-adic integers? In the former case, there is a trivial subact, while in the latter the action in simple. Take the case $x=1\in \mathbb{Z}_p$ and ask yourself when can you have $nx = x$ depending on $n$ (i.e. trivial action of $n$)? – Kolja Oct 19 '22 at 14:21
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    @Kolja Thanks very much, Here $\mathbb{Z}_p$ is just thought as a set (not a module). We have to show that if $A$ is a sub-act of $\mathbb{Z}_p$, then $A$ is either ${ 0}$ or $\mathbb{Z}_p$. – M.Ramana Oct 19 '22 at 14:27
  • Only a comment, but isn't it a bit weird to tag this question as 'calculus' (or 'modules' and 'group theory' for that matter)? – peter a g Oct 19 '22 at 15:34
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    Why are you excluding 0 when looking for a sub-act? There did not seem to be anything about that in the definition. – Tobias Kildetoft Oct 20 '22 at 09:19
  • I agree with @TobiasKildetoft , $A$ is just a set, so there is no precedence of $0$ with respect to other elements. It would make sense if you give $A$ some structure and expect it to have a certain relation with respect to the action, like with ring modules. – Kolja Oct 20 '22 at 09:28

1 Answers1

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As you noticed correctly, it is enough to show that for all $x\not\equiv 0 \bmod p$ there exists an $n_0\in \mathbb{N}$ such that $n_0 x \equiv 1 \bmod p$.

This follows directly from $\gcd(x, p) = 1$ whenever $x \not\equiv 0 \bmod p$.

For more information see https://en.wikipedia.org/wiki/Modular_multiplicative_inverse or, for example, If $q$ is coprime to $a$ then $a\mid nq-1,\,$ so $q$ is invertible mod $a$ .

Kolja
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  • Thank you so much for your help. I appreciate it. – M.Ramana Oct 20 '22 at 08:35
  • Please note comment of @TobiasKildetoft, my answer was provided given your assumption that ${0}$ is ignored as a trivial sub-act. However, it doesn't make much sense to exclude ${0}$ because $A$ is only a set without any structure, so excluding ${0}$ makes as much sense as excluding ${1}$. There IS a sub-act, but make sure that your definitions of simple acts align with this exclusion. – Kolja Oct 20 '22 at 09:31