# Functions in modular arithmetic that are injective, surjective, or invertible.

There is a question I am struggling on.

Let f : Z12 → Z12 : x ↦→ 9 x + 1 where arithmetic is done modulo 12.

(a) Show that f is neither injective, nor surjective.

(b) Now consider g, where g : Z12 → Z12 : x ↦ 7 x + 1. Show that g is invertible. Calculate g^−1(0) and g^−1(11). Find a formula for g^−1(x).

Hint: first find a number x so that 7 x is equivalent to 1 modulo 12. Now use this judiciously.

Can someone tell me if my working out is correct and how to improve the answer.

Here is what I worked out : a) f is not injective, because f(0) = 1 = f(4), but 1 and 4 are distinct mod 12.

Next, f is not surjective, because there is no x in Z12 such that f(x) = 9x + 1 = 2 (mod 12), since this is equivalent to 9x = 1 (mod 12) and 9 is not invertible mod 12, because gcd(9, 12) = 3 > 1.

(b) Given y in Z12, we want to find y in Z12 such that y = g^(-1)(x). <==> g(y) = x. <==> 7y + 1 = x (mod 12) <==> 7y = x - 1 (mod 12) <==> 7 * 7y = 7(x - 1) (mod 12) <==> y = 7x - 7 (mod 12), since 49 = 1 (mod 12) <==> y = 7x + 5 (mod 12), since -7 = 5 (mod 12).

That is, g^(-1)(x) = 7x + 5.

In particular, g^(-1)(0) = 5, and g^(-1)(11) = 82 = 10 (mod 12).

• In the title, do you mean surjective Oct 8 '16 at 8:12
• yes sorry that was a typo Oct 8 '16 at 8:18

You could shorten you answer observing that, since $\mathbf Z/12\mathbf Z$ is a finite set, injective $\iff$ surjective$\iff$ bijective.
Furthermore, in any ring $R$, the map $x\mapsto ax+b$ is injective (resp. surjective, bijective) if and only if $x\mapsto ax$ is, and: