I have a question about the automorphism group of the cyclic group of order $p^2$ where $p$ is a prime. In my notes, I have written that this group has a normal Sylow $p$-subgroup (i.e. only one Sylow $p$-subgroup). However, now when I am re-reading my notes, I cannot exactly see why. I know that $\vert \text{Aut}(C_{p^2})\vert = p(p-1)$ using Euler's totient function, so it is not because $\text{Aut}(C_{p^2})$ is a $p$-group. Did I write down something wrong or can somebody tell me why $\text{Aut}(C_{p^2})$ has a normal Sylow $p$-subgroup?


That subgroup is normal because the automorphism group is abelian. In fact, if $p$ is odd, then the automorphism group is cyclic.

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    $\begingroup$ For $p^2$, it works for $p=2$ too. $p^3$ is the first time $p=2$ is weird. $\endgroup$ May 25 '14 at 18:37
  • $\begingroup$ Oh, right, I forgot that. $\endgroup$
    – Nishant
    May 25 '14 at 18:39
  • $\begingroup$ Of course! Thank you @Nishant $\endgroup$
    – user129954
    May 25 '14 at 18:43

It is because

  • $p-1$ does not have a factor of $p$
  • $n_p=1+kp>p$ for $k\ge 1$
  • $n_p=1$ implies ????

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