Possibilities of an automorphism on $Z_{20}$

Question

This is a question from the book Contemporary Abstract Algebra, specifically chapter 6 on isomorphisms. The question is:

Suppose that $\phi: Z_{20} \to Z_{20}$ is an automorphism and $\phi(5) = 5$. What are the possibilities of $\phi(x)$?

Attempt of Solution

I have made an attempt at solving it. Firstly, we know that an isomorphism carries the identity of the first group to the identity of the second group. Since the identity is $0$, and it is an automorphism, so $\phi(0) = 0$.

Next up, we know that an isomorphism preserves the order of an element. So I made a list of all elements of $Z_{20}$ with the same order. For reference, here's the list (excluding $5$ and $0$): \begin{align*} 1) & \, \{1, 3, 7, 9, 11, 13, 17, 19 \} &\text{ (Order is 20) } \\ 2) & \, \{2, 6, 14, 18 \} &\text{ (Order is 10) } \\ 3) & \, \{4, 8, 12, 16 \} &\text{ (Order is 5) } \\ 4) & \, \{10 \} &\text{ (Order is 2) } \\ 5) & \, \{15 \} &\text{ (Order is 4) } \\ \end{align*} So we can restrict $\phi$ in that if it is an automorphism, it must map $0, 5, 10$ and $15$ to themselves. Also, it can only map the rest of the elements to some element in their set otherwise order will change.

That give $8! \cdot 4! \cdot 4!$ possibilities in total. Is this approach correct, or I'm missing something else?

Answer 1

Expanding the comment, the only possible automorphisms are the ones where $\phi(1)$ are sent to another element of order 20. Once you have that, everything else is completely determined by the homomorphism property. So all you have to do is check which of the elements of order 20 that when you multiply by 5, mod 20, give you 5. Those are the ones which preserve $\phi(5)=5$

Answer 2

The number of automorphisms from $\mathbb{Z_{n}}\rightarrow\mathbb{Z_{n}}$ is equal to $\phi(n)$ where $\phi$ denotes the Euler's totient function. You just have to think like this:- since $1$ generates $\mathbb{Z_{n}}$ , and $f$ is an automorphism, $f(1)$ must also generate $\mathbb{Z_{n}}$ . So you just have to map $1$ to some generator and it would give you the different automorphisms. Now $\mathbb{Z_{n}}$ has $\phi(n)$ generators. So you get your answer. Also the Automorphism group of $\mathbb{Z_{n}}$ will be isomorphic to $\mathbb{Z^{\times}_{n}}$ or $U_{n}$ where $U_{n}$ or $\mathbb{Z^{\times}_{n}}$ denotes the multiplicative group of units in $\mathbb{Z_{n}}$. $U_{n}$ has order $\phi(n)$ as only the elements which are coprime to $n$ are the units and number of positive integers less than $n$ and coprime to $n$ is precisely the definition of $\phi(n)$.

In particular $\mathbb{Z_{20}}$ has $\phi(20)=\phi(4)\cdot\phi(5)=2\cdot 4 =8$ automorphisms.

Answer 3

Let $x=\phi(1)$ so that $\phi(5)=5x$. Then the automorphisms you want are those corresponding to the solutions of the congruence $5x\equiv5\bmod 20$ which are coprime to $20$. The previous congruence is equivalent to $$ x\equiv1\bmod 4 $$ and thus $x\in\{1,9,13,17\}$