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So on this webpage: https://crypto.stanford.edu/pbc/notes/numbertheory/gen.html I read that

Let p be prime, ${Z_{P}}^{\times}$ contains exactly $\phi (p-1)$ generators.

Additionally, in this post:Order of automorphism group of cyclic group I read that

Aut()|=() where ()is Euler's function. (G is a cyclic group of order m)

The point where Im confused is that let Q be a normal sylow q group of a group G of order pq ($p < q$). Then we have that $Aut(Q) \cong (Z/qZ)^{\times} \cong Z/(q-1)Z$. But for cyclic groups to be isomorphic they need to have the same order and the same number of generators. The order of the groups is q-1. This is obvious. But according to the websites above, $(Z/qZ)^{\times}$ has $\phi (q-1)$ generators, and the $Aut(Q)$ has $\phi(q)$ generators (because its enough to know how many generators $Q$ has). I do not see how $\phi(q-1) = \phi(q)$ and therefore how the first two groups have the same amount of generators? I see how they all have the same order. This is extremely confusing, any help would be appreciated.

froitmi
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    $\text{Aut}(Q)$ does not have $\phi(q)$ generators. It's $Q$ that has $\phi(q)$ generators, so there's no contradiction. – Qiaochu Yuan Jun 27 '24 at 20:29
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    Ah, and the second quoted statement says that $\text{Aut}(Q)$ has $\phi(q)$ elements, not generators. – Qiaochu Yuan Jun 27 '24 at 20:44
  • Oh yes I see that now, i just got really confused, its my first time seeing the function and I have no one to talk to about math so I tend to confuse myself. I was getting the order and generators all mixed up. Thank you. – froitmi Jun 27 '24 at 20:46

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