Homomorphism from $\mathbb{Z}_p \to \mathrm{Aut}(\mathbb{Z}_p)$ for $p$ prime

Question

I am trying to get all homomorphisms from $\mathbb{Z}_p \to \mathrm{Aut}(\mathbb{Z}_p)$ for a prime number $p$. I know there should only be one, the trivial one since $\mathrm{Aut}(\mathbb{Z}_p) \cong \mathbb{Z}_{p-1}$. I understand that for $ p=2$ we can only have one homomorphism since we only have one automorphism from $\mathbb{Z}_2 \to \mathbb{Z}_2$. However for $ p=3$ I found two potential homomorphisms: $ \phi : \mathbb{Z}_3 \to \mathrm{Aut}(\mathbb{Z}_3)$ such that: $$\begin{align*} 0 \to \varphi_1 &:= \varphi (1) = 1;\\ 1 \to \varphi_1 &:= \varphi (1) = 1;\\ 2 \to \varphi_1 &:= \varphi (1) = 1 \end{align*}$$

However I do not know why: $ \phi : \mathbb{Z}_3 \to \mathrm{Aut}(\mathbb{Z}_3)$ such that: $$\begin{align*} 0 \to \varphi_1 &:= \varphi (1) = 1;\\ 1 \to \varphi_2 &:= \varphi (1) = 2;\\ 2 \to \varphi_1 &:= \varphi (1) = 1 \end{align*}$$

should not work, since we have $\phi(1) = \phi(0)\phi(1) = \varphi_1\varphi_2$ which works when plugging in the values $0,1,2$. Furthermore, $\phi(2) = \phi(1)\phi(1) = \varphi_2\varphi_2$ with $ \varphi_2\varphi_2(1) = \varphi_2(2)= 1 = \varphi_1(1)$ and $ \varphi_2\varphi_2(2) = \varphi_2(1)= 2 = \varphi_1(2)$. Where is my mistake and how would the argument that there is only one homomorphism from $\mathbb{Z}_p \to \mathrm{Aut}(\mathbb{Z}_p)$ generalize for higher primes? Is there a way without using $ \mathrm{Aut}(\mathbb{Z}_p) \cong \mathbb{Z}_{p-1}$ and rather proving it directly? Any help is much appreciated.

Answer 1

The $\phi$ you wrote down is not a homomorphism. Compare $\phi(1+2)(1) = \phi(0)(1) = \varphi_1(1) = 1$ with $(\phi(1) \circ \phi(2))(1) = \varphi_2(\varphi_1(1)) = \varphi_2(1) = 2$.

Answer 2

I think I've a proof that there can only exist a trivial homomorphism between $ \mathbb{Z}_p $ and $ Aut(\mathbb{Z}_p)$. I would be glad if someone could confirm it:

Let $\phi : \mathbb{Z}_p \to Aut(\mathbb{Z}_p) $. Then necessarily we have $\phi(0)=id$. Let $\varphi_a$ be defined as the automorphism $\varphi$ such that $ \varphi(1) = a$. Since $1$ is a generator of $\mathbb{Z}_p $ this uniquely defines the automorphism for $ a \ne 0$. Especially $\varphi_1 = id$.

Now consider any homomorphism $\phi$ that maps $ 0 \to \varphi_1 $ and $ 1 \to \varphi_a $ such that $ a \in \mathbb{Z}_p$ and $a \ne 0$.

It follows that $\phi(k) = (\phi(1))^k$ under composition. Thus especially $ (\phi(1))^k = (\varphi_a)^k$. Furthermore, $\phi(k)(1) = (\phi(1))^k(1) = a^k$. Since $\varphi_a(a) = a*\varphi_a(1) = a^2$ then the result follows via induction.

Chosing $k=p$ we obtain: $\phi(0)(1)=\phi(p)(1)=a^p$. From Fermat's little theorem we conclude that $a^p = a$ in $\mathbb{Z}_p$ and therefore, $\phi(0)(1)=\phi(p)(1)=a^p=a$. However, since $\phi(0) = id$ we need $ a=1$ for $\phi$ to be a homomorphism. Since the homomorphism is uniquely determined by the value of $1$ we at most have one choice for the homomorphism, namely the one that maps every element to $\varphi_1 $. One verifies easily that this is an homomorphism, since $\phi(a+b) = \varphi_1 = id = id^2 = \phi(a) \phi(b) $.