Let $\phi:\Bbb{Z}_{20} \to \Bbb{Z}_{20}$ be an automorphism and $\phi(5)=5$. What are the possibilities of $\phi(x)$?

Question

Same question here: Possibilities for $\phi(x)$, but I'm striving for a more general solution. Please tell me what I'm doing wrong or if my reasoning is correct.

Scratch/solution: $\phi$ is an isomorphism so $\phi(0)=0$ and every generator of $\Bbb{Z}_{20}$ is mapped to a generator in $\Bbb{Z}_{20}$. Observation: We want the following properties \begin{align} \phi(5)&=\phi(1+1+1+1+1)=5\phi(1)=5 \pmod{20}\\ \phi(5)&=\phi(\sum_{i=1}^{15}3)=\sum_{i=1}^{15}\phi(3)=15\phi(3)=5 \pmod{20}\\ \vdots\\ \text{If} \quad \phi(3)&=1, \text{then}\quad \phi(5)=15\phi(3)=15 \pmod{20}\\ \text{If} \quad \phi(3)&=11, \text{then}\quad \phi(5)=15\phi(3)=15\times 11=165=5\pmod{20} \end{align} So I became inclined to find an $\alpha$ such that $15\alpha \equiv5\pmod{20}$, this is in the case of finding a possible image of $\phi(3)$; that is to let $\phi(3)=\alpha$. In general, I noted that for all generators $\gamma\in\Bbb{Z}_{20}$, $\exists k\in\Bbb{Z}_20$ such that $x\equiv \gamma k\pmod{20}$ where $x$ is that arbitrary element of $\Bbb{Z}_{20}$. Which implies $\phi(x)=\phi(\gamma k)=k\phi(\gamma)\pmod{20}$. By finding all $y$ such that $ky \equiv \pmod{20}$ we can let $\phi(\gamma)=y$ and call $\phi(x)=k\phi(y)=ky\pmod{20}$ a possible image of $\phi(x)$.

Answer 1

It's a lot easier to see exactly how flexible the automorphisms are (and count them) if you use $\Bbb Z_{20}\cong \Bbb Z_4\times\Bbb Z_5$, with $5\mapsto (1, 0)$

This is known as the Chinese remainder theorem: If $m, n$ are coprime integers, then $\Bbb Z_{mn}\cong \Bbb Z_m\times \Bbb Z_n$. The most natural isomorphism is $1\mapsto (1, 1)$, so that's the one I'll use here. this gives the $5\mapsto(1, 0)$ above.

An automorphism of $\Bbb Z_4\times \Bbb Z_5$ must send $(1, 0)$ to an element of order $4$ (i.e. $(1, 0)$ or $(3, 0)$) and it must send $(0,1)$ to an element of order $5$ (i.e. $(0,i)$ for $1\leq i\leq 4$), and any automorphism is uniquely determined by where it sends these two elements.

We want all automorphisms which fix $5\in \Bbb Z_{20}$. That corresponds to automorphisms which fix $(1, 0)$ in $\Bbb Z_4\times\Bbb Z_5$. That means the only leeway we have is where $(0,1)$ is sent. We have four options, and they all work. Thus the four maps we are after all map $(1,0)$ to $(1,0)$, and then map $(0,1)$ to either one of the four order-$5$ elements $(0,i)$.

In order to translate back to $\Bbb Z_{20}$ I think it's easiest to see what happens to $(1, 1)$: It is sent to some element $(1, i)$ where $1\leq i\leq 4$. Any such choice gives a valid automorphism. Taking our designated isomorphism back to $\Bbb Z_{20}$, we get the following correspondences between automorphisms: $$ \begin{array}{|c|c|} \hline \text{image of (1, 1)}&\text{image of $1$}\\ \hline (1, 1) & 1\\ (1, 2) & 17\\ (1, 3) & 13\\ (1, 4) & 9\\\hline \end{array} $$

Answer 2

$$5\phi(1) \equiv 5 \pmod{20} \iff \phi(1)\equiv 1 \pmod4 \iff \phi(1)\equiv x \pmod{20} $$ where $x\in \{1, 5, 9, 13, 17\}$. This gives us $5$ automorphisms, each of the form $$\phi(n) = nx $$

But since $5$ is not a generator of $Z_{20}$ hence $\phi(1)\neq 5$ Hence, $\phi(x)=x,9x,13x,17x$ are the only possibilities.

Answer 3

We know, an isomorphism maps generators to generators. Gen Z_20 = {1,3,7,9,11,13,17,19}. These are the possible competitors for automorphisms. The other condition according to question is Φ(5)=5, say this condition is (C1). To be more accurate Φ(5)=5(mod 20). Now, Φ(x)=x satisfies C1. Φ(x)=3x will give Φ(5)=15(mod 20) doesn't satisfies C1. Φ(x)=7x wil give Φ(5)=15(mod 20) doesn't satisfies C1, and so on. Eventually we get the automorphisms as x, 9x, 13x, and 17x.