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I want to show that $\text{Aut}(\mathbb{Z}_p)$ is isomorphic to $\mathbb{Z}_{p-1}$.

The group $\mathbb{Z}_p$ is cylcic and is generated by one element. The possible generators are $\{1,2,\dots , p-1\}$.

Each automorphism maps each of the elements of the set $\{1,2,\dots , p-1\}$ to one of the elements $\{1,2,\dots , p-1\}$.

So, we have the function $f_m(x)=mx\pmod p, \ 1\leq m\leq p-1$, right?

To show that $\text{Aut}(\mathbb{Z}_p)$ is isomorphic to $\mathbb{Z}_{p-1}$ do we define the function $$h:\text{Aut}(\mathbb{Z}_p)\rightarrow \mathbb{Z}_{p-1} \text{ with } \\ f_m\mapsto (m-1)\pmod {p-1} , \ 1\leq m\leq p-1$$ ?

Mary Star
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  • You'll probably have an easier time if you define $h:\mathbb Z_p\to\text{Aut}(\mathbb Z_p)$ as $h(m)=f_m$. – Tim Raczkowski Feb 18 '16 at 21:47
  • You mean $h:\mathbb Z_{p-1}\to\text{Aut}(\mathbb Z_p)$ ? @TimRaczkowski – Mary Star Feb 18 '16 at 21:53
  • Yes. Sorry, that was a typo. – Tim Raczkowski Feb 18 '16 at 21:54
  • It is the same what I proposed it's only the reverse function, or not? @TimRaczkowski – Mary Star Feb 18 '16 at 21:55
  • Yes. That's right. – Tim Raczkowski Feb 18 '16 at 21:55
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    The same question was asked here with good answers. – Dietrich Burde Feb 18 '16 at 21:58
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    @MaryStar You are confusing $\mathbb{Z}p^{\times}$ (a multiplicative group) with $\mathbb{Z}{p-1}$ (an additive group). They are isomorphic, but not the same thing. – David Hill Feb 18 '16 at 22:00
  • So the functions that I wrote would be a mapping to $\mathbb{Z}p^{\times}$ and not to $\mathbb{Z}{p-1}$ ? @DavidHill – Mary Star Feb 18 '16 at 22:11
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    @MaryStar that is correct. – David Hill Feb 18 '16 at 22:23
  • What could I change so that it is a mapping to $\mathbb{Z}_{p-1}$ ? @DavidHill – Mary Star Feb 18 '16 at 22:31
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    Well, you need to compose with an isomorphism $\mathbb{Z}p^{\times}\to\mathbb{Z}{p-1}$. That means you have to choose a generator $[m]\in\mathbb{Z}p^\times$ and every element in $\mathbb{Z}_p^\times$ can then be written as $[m^k]$ ($0\leq k\leq p-2$). Then, your map is $f{m^k}\mapsto k$. – David Hill Feb 18 '16 at 22:43
  • Is $\mathbb{Z}p^\times$ cyclic? $$$$ So we have $$h_1:\text{Aut}(\mathbb{Z}_p)\rightarrow \mathbb{Z}_p^\times \text{ with } \ f{m+1}\mapsto m\pmod {p-1} , \ 0\leq m\leq p-2$$ right? $$$$ Why do we take for $\mathbb{Z}p^{\times}\to\mathbb{Z}{p-1}$ the map $f_{m^k}\mapsto k$ ? Isn't $f_{m^k}$ an element of $\text{Aut}(\mathbb{Z}_p)$ ? @DavidHill – Mary Star Feb 22 '16 at 14:35
  • Is a composition necessary? Could we show that the two groups are isomorphic to each other without the composition? @DavidHill – Mary Star Feb 23 '16 at 00:15

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