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I'm looking for an example of a prime $p$ and an abelian group $G$ of order $p-1$, such that:

  1. $G$ has a subgroup $H\cong C_q\times C_q$, where $q$ is another prime (of course, $q^2\mid p-1$);
  2. $G$ acts by automorphisms transitively on $C_p\setminus\{0\}$.

Is there any? Or are 1 and 2 mutually excluding for every $p$?

Kan't
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1 Answers1

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First note the automorphism group of $C_p$ (which is the same as the automorphism group of $C_p \setminus \{ 0 \}$, since every automorphism fixes $0$) is isomorphic to $C_{p-1}$ (see here, for instance).

Now if $G$ satisfies condition (2) and acts transitively on $C_p \setminus \{ 0 \}$, then by the orbit-stabilizer theorem and the fact that $|G| = |C_p \setminus \{0\}| = p-1$ we learn that the stabilizer of every $k \in C_p \setminus \{0\}$ has to be trivial. In particular, $G$ injects into $C_{p-1} = \text{Aut}(C_p \setminus \{0\})$, so for cardinality reasons again we learn $G \cong C_{p-1}$.

So the question is whether $C_{p-1}$ can satisfy (1) and have a subgroup isomorphic to $C_q \times C_q$, but of course it can't since every subgroup of a cyclic group is cyclic (see here), and $C_q \times C_q$ isn't.

I hope this helps ^_^

Kan't
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Chris Grossack
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  • Thank you. I should have mentioned explicitly that, in this context, I "don't know" that the automorphism group of a cyclic group of prime order is cyclic. Moreover, I'm curious about the writing "$Aut(C_p\setminus{0})$", as what is in round brackets is not a group. – Kan't Mar 30 '25 at 08:16
  • But your question has now been answered. There are no such examples. – Derek Holt Mar 30 '25 at 09:43