Prove for any $[a] \in \mathbb{Z}^\times_n$ multiplication by $[a]$ defines an automorphism $\alpha_{[a]}: \mathbb{Z}_n \to \mathbb{Z}_n$

Question

For this we need to show that map $\alpha_{[a]}: \mathbb{Z}_n \to \mathbb{Z}_n$ defined by $\alpha_{[a]}([k]) = [ak]$ is a group isomorphism; that is, a bijective homomorphism. I understand what's required I think I'm just getting a little confused about which elements to use to check the properties specifically. My intuition tells me to check, say, the homomorphism property with the elements $[a],[b] \in \mathbb{Z}_n^\times$ but when I tried that I got: $$\alpha_{[ab]}([k]) = [abk] = [a][bk] = ...$$ where I would imagine we're looking for a product of the form $$[ak][bk] = \alpha_{[a]}([k])\alpha_{[b]}([k])$$ However, if we were to carry out the homomorphism property regarding two elements in the domain $[k],[q] \in \mathbb{Z}_n$ we would get $$\alpha_{[a]}([k+q]) = [a(k+q)] = [ak+aq] = [ak] + [aq] = \alpha_{[a]}([k])\alpha_{[a]}([q])$$ which seems like the homomorphism property is satisfied.

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So my main question here is which one is correct? Should I be checking these properties (homomorphism, injectivity and surjectivity) against the elements in $\mathbb{Z}_n^\times$ (the subscripts) or the elements of the domain $\mathbb{Z}_n$ (the arguments). And how should I go about understanding which one to check various properties against going forward?

Answer 1

Fix $[a]\in\Bbb Z_n^\times$. You need to show the homomorphism property

$$\alpha_{[a]}([k]+[l])=\alpha_{[a]}([k])+\alpha_{[a]}([l])$$

for all $[k],[l]\in\Bbb Z_n$, together with either surjectivity or injectivity (which are each sufficient to show bijectivity; and bijectivity only holds for $[a]\in\Bbb Z_n^\times$). You also need to show $\alpha_{[a]}$ is well-defined, which means that for all $[x],[y]\in\Bbb Z_n$,

$$[x]=[y]\implies \alpha_{[a]}([x])=\alpha_{[a]}([y]).$$

Answer 2

I would suggest that you don't bother with the specific case of $\mathbb Z_n$. Just prove that for any ring $(R, + , \cdot)$ and $a \in R$ the map

$$\varphi_a: x \mapsto ax$$ is a homomorphism of the group $(R, +)$. That is easy and will avoid cumbersome notations.

Then you have to prove that for $[a] \in \mathbb Z_n^\times$, $\alpha_{[a]}$ is bijective. And that is quite easy as by definition, it exists $[b] \in \mathbb Z_n^\times$ such that $[b][a]= [1]$ and therefore $\alpha_{[b]}$ is the inverse of $\alpha_{[a]}$.