# A proof about Automorphism in congruence class

Suppose $$gcd(m,n)=1$$, and let $$F :Z_n→Z_n$$ be defined by $$F([a])=m[a]$$. Prove that $$F$$ is an automorphism of the additive group $$Z_n$$. I find it is diffcult to prove $$F$$ is injective and surjective. Could you please to help me proof it with all the details. I type it roughly, and i am sorry and sincerely looking for a result.

• Is it clear to you, that the equation $ma=0 \ mod(n)$ has trivial solution? In other words, that $\ker(F) = 0$ and hence the map is injective? An injective endomorphism of a group is automaticaly an isomorphism. – Lada Dudnikova Apr 22 at 8:14
• @LadaDudnikova "An injective endomorphism of a group is automatically an isomorphism" - this is not true, see here. – Dietrich Burde Apr 22 at 10:48
• @Lada Dudnikova Maybe for the finite group,then "An injective endomorphism of the group to itself is automatically an isomorphism" is true. However, could you please show me that how can i use $gcd(m,n)=1$, and the trivial solution to obtain the map is injective. I am confused about this corollary. – B1s Apr 22 at 11:03
• @DietrichBurde sorry, forgot to mention finiteness of the group order here. – Lada Dudnikova Apr 22 at 11:03

## 2 Answers

To show that the map is injective and surjective is equivalent to showing that the map has a two-sided inverse.

The extended Euclidean algorithm yields that there are numbers $$m', n'$$ such that $$m m' + n n '= 1.$$ Consider the map $$G : Z_{n} \to Z_{n}$$ given by $$G([b]) = m' [b]$$. Then for all $$a$$ one has $$G \circ F([a]) = G(F([a])) = G(m [a]) = m'm [a] = (1 - n' n) [a] = [a],$$ as $$n [a] = $$. Similarly $$F \circ G([b]) = [b]$$ for all $$b$$.

I shall prove two things

1. The solution of $$am = 0 \ (mod \ n)$$ is trivial
2. (1) means that the map is injective.

Let there a non-zero element s.t. $$am = 0$$ modulo $$n$$. The multiplication by integer in the group means consequtive addition(substraction) of the element to itself. Knowing that $$gcd(m,n) = 1$$ means that there is such a pair $$(x',y')$$ that $$x'm+y'n = 1 \implies x'm = 1$$ modulo $$n$$

This means $$1 = a(x'm) = x'(am) = 0$$ contradiction.

Let the injectivity be violated. Then there is $$ma = mb = c \neq 0$$ modulo $$n$$

that means $$m(a-b) = c \neq 0$$ modulo $$n$$ which is a contradiction with the previous statement.

I admit that I operate rather dirty making commutative permutations and not justifying why in cyclic group I can do it. Do you need it to be justfied?