Prove that the only automorphism of order 2 of $\mathbb{Z}_q$ is $m \mapsto -m$ for $q = 3$ and $q = 5$ and $q = 7$

Question

My assignment asks me to prove that the only automorphism of order 2 of $\mathbb{Z}_q$ is $m \mapsto -m$ for $q = 3$ and $q = 5$ and $q = 7$. I have been stuck for ages, and now I wonder if it is true. I need some help to get started. This is my attempt: Assume $k$ and $q$ are relative prime (this is necessary otherwise $\phi$ is not an automorphism), then: $$ m = \phi^2(m) = k^2m \implies k^2 \equiv 1 \mod q $$ But I don't know how to go on from there. I cant see that $k = -1$ is the only option.

Answer 1

Hint: When $q$ is prime, $ k^2 \equiv 1 \bmod q \iff k \equiv \pm1 \bmod q$. What is the order of $x \mapsto kx$ when $k=1$ ?

Answer 2

Suppose $k^2\equiv 1 \bmod p$, then $(k-1)(k+1)\equiv 0 \bmod p$. That is the reason why there are only $2$ options.

In general we can do these sort of things in any integer domain.( polynomials cant have more roots than their degree)

Answer 3

For $q$ prime, $\mathbb{Z}_q$ is a field, so $k^2-1=0$ there has only two solutions, $k=1$ or $k=-1$.

Answer 4

You are almost there.

You're right that $k=-1$ is a possibility, since $k^2\equiv_q 1$ (you also know that there are atmost two values $k$ can have, since $\mathbb{Z}_q$ is a field). Remember that if $k=1$, i.e, $m\mapsto m$ then the order of the automorphism is $1$, not $2$.