Let $f: Z_{12} \to Z_{12}$ be defined by $f(x)=5x+8$. Prove that $f$ is injective and surjective but not an isomorphism.

Question

I already know how to prove this is an injection, but showing the function is onto is proving to be tricky. I think. If I want to show that for all $b$ in $Z_{12}$ there exists an $a$ in $Z_{12}$ such that $f(a)=b$. Wouldnt I want to say that $a=(b-8)/5$ is in $Z_{12}$? But it's not true unless that is under mod $12$. I need a little bit of help please.

Answer 1

You said you already have injectivity.

The comments provide two ways to prove surjectivity. Either argue that an injective map from a set of 12 elements to a set of 12 elements has to be surjective or compute explicitely in $\mathbb{Z}_{12}$ to show it directly.

The final part aks to prove that it is not an isomorphism. The trick here is that if one writes $\mathbb{Z}_{12}$ one usually thinks of it as a group not just a set. Your map $f$ is an isomorphism of sets (it has to be because the sets are finite and it is injective and surjective) but it is not an isomorphism of groups. An isomorphism of groups needs to preserve the group structure as well. This implies that the identity element is mapped to the identity, this is not the case here.

Answer 2

If $\bar x,\bar y\in \mathbb Z/n\mathbb Z$ with $f(\bar x)=f(\bar y)$, then $5\bar x+\bar 8 = 5\bar y + \bar 8$, from which it is immediate that $\bar x=\bar y$ (division by $5$ is permissible as $5$ and $12$ are coprime). Now let $\bar z\in\mathbb Z/n\mathbb Z$. Then because $5$ and $12$ are coprime, there exists $\bar w\in\mathbb Z/n\mathbb Z$ such that $\bar z = \bar 5\bar w$. It follows then that $f(\bar w-\bar 8)=\bar z$.

To see that $f$ is not a homomorphism, consider that $f(\bar 0) = \bar 5\bar 0 + \bar 8=\bar8\ne\bar 0$, so $f$ does not map the identity to the identity.

As a side note, we are considering multiplication as well as addition in $\mathbb Z/n\mathbb Z$. So shouldn't this be a question of rings and not a question of groups?