Prove that α is an automorphism of $\mathbb{Z}_n \to \mathbb{Z}_n$.

Question

Let $n=2k+1$ be odd. Prove that the mapping $\alpha : \mathbb{Z}_n \rightarrow \mathbb{Z}_n$ given by $\alpha (s) = 2s$ (mod $n$) for every $s \in \mathbb{Z}_n$ is an automorphism from $\mathbb{Z}_n$ to $\mathbb{Z}_n$.

I am struggling at answering this question and would love some tips.

I managed to solve the mapping, 1-1 and operation preserving parts. But not the onto.

This is what I have so far

First, we define a candidate for the isomorphism. Let $\alpha : \mathbb{Z}_n \rightarrow \mathbb{Z}_n$ be an isomorphism given by the following map: $\alpha(s)=2s$ (mod $n$). Now we need to prove its one-to-one. Assume that $\alpha(a_1) = \alpha(a_2)$ for some $a_1$,$a_2 \in \mathbb{Z}_n$. Note that $\alpha (a_1) = 2a_1$ (mod $n$) and $\alpha(a_2) = 2a_2 $ (mod $n$). So $\alpha(a_1)=\alpha(a_2) \Rightarrow 2a_1 = 2a_2$ (mod $n$). Notice here since $2a_1 $ (mod $n$) $\in \mathbb{Z}_n$ and $2a_2$ (mod $n$) $\in \mathbb{Z}_n$, then cancellation law holds. Therefore $a_1$ = $a_2$.

Answer 1

Note that $2a(k+1)=2ak+a+a = (2k+1)a +a \equiv a\pmod n$.

That is, given any $a$, multiplying by $2$, followed by multiplying by $k+1$ gives back $a$. So multiplying by $k+1$ is the inverse operation for the given automorphism. This shows it is onto.

Answer 2

There are a finite number of elements in $\Bbb Z_n$, and you've shown that $\alpha$ is 1-1. That implies that it's onto.

For a more direct proof, given a congruency class $[t]\in \Bbb Z_n$, there is some integer $t\in \Bbb Z$ representing it. If $t$ is even, then $\alpha([t/2])=[t]$. If $t$ is odd, then $t+n$ is even, $[t]=[t+n]$, and $\alpha([(t+n)/2])=[t+n]=[t]$.

So any congruency class is in the image of $\alpha$, and therefore $\alpha$ is onto.