Regular action and integers modulo $p-1$ and $p$

Question

For $p$ a prime number, let's denote:

$$G:=\mathbb Z/(p-1)\mathbb Z = \{[i]_{p-1}\mid i=0,1,\dots,p-2\}$$

and:

$$X:=\mathbb Z/p\mathbb Z\setminus \{[0]_p\} = \{[j]_p\mid j=1,2,\dots,p-1\}$$

Does $G$ regularly (=transitively+freely) act on $X$, somehow?

My first idea was to try with $[i]_{p-1}\mapsto([j]_p\mapsto[i+j]_p)$, but of course it can't work, as $[i+j]_p=[0]_p$ as soon as $i+j=p$, and hence $[i+j]_p$ may well not even lie in $X$.

Edit. In the spirit of the question, I "don't know" that $(\mathbb Z/p\mathbb Z)^\times$ is cyclic (of order $p-1$).

The group $X=(\Bbb Z/p\Bbb Z)^$ is isomorphic* to $G$. The action by translations of any group on itself is transitive and free. — Anne Bauval, Jan 04 '24 at 14:11
Does this answer your question? Show that the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^*$ is cyclic (and every cyclic group $G$ of order $n$ is naturally isomorphic to $(\Bbb Z/n\Bbb Z,+)$, once you choose a generator of $G$) — Anne Bauval, Jan 04 '24 at 14:23
Is there any chance to deal directly with $G$ for what it is, namely the additive group of the integers modulo $p-1$, @AnneBauval? — citadel, Jan 04 '24 at 14:26
The answer to the question that you asked is yes because $|G| = |X|$. The result follows from Cayley's Theorem. The structure of $({\mathbb Z}/p{\mathbb Z})^*$ is irrelevant. — Derek Holt, Jan 04 '24 at 18:17
Sure! $G\stackrel{Cayley}{\hookrightarrow}S_G\stackrel{|G|=|X|}{\cong} S_X$. And Cayley's embedding is (equivalent to) a regular action. I'll try to build them (composition) up explicitly. Thank you, @DerekHolt. — citadel, Jan 04 '24 at 18:49

Shaun · Answer 1 · 2024-01-04T14:26:43.677

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Each group acts on itself transitively and freely, and $\Bbb Z/(p-1)\Bbb Z\cong\Bbb Z_p^*$.

Note: This is CW. Credit to @AnneBauval. See the comments.

edited Jan 04 '24 at 14:26

answered Jan 04 '24 at 14:21

Shaun

44,997

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(+1 to CW) Maybe you were right to edit an answer. I was intending to link instead as a duplicate, but didn't know which one of the 2 parts of the argument was less known to the OP. – Anne Bauval Jan 04 '24 at 14:29
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Note that as I pointed out in a comment, you don't actually need the fact that $Z/(p-1)Z$ and $Z_p^*$ are isomorphic to answer the question. It is enough that they have the same order. – Derek Holt Jan 04 '24 at 19:35
Why to mention $Z_p^*$ at all, @DerekHolt? – citadel Jan 04 '24 at 20:49
Because you mentioned it yourself in the question. – Derek Holt Jan 04 '24 at 22:45
Not originally, only later when others' answer and comments brought it into the discussion. – citadel Jan 05 '24 at 06:02

Regular action and integers modulo $p-1$ and $p$

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