Show that, for every $x\ \epsilon \ C_{m}$, we have that $ord(f(x))$ is a divisor of d.

Question

Let $C_{m},C_{n}$ be two finite cyclic groups of cardinality $ n,m \ \epsilon\ \mathbb{N}$, where $\mathbb{N}$ is defined without $0$. Let $d=g.c.d(m,n)$ be the greatest common divisor of $n$ and $m$. We want to classify all morphism of groups $f:C_{m}\rightarrow C_{n} $

1) Show that, for every $x\ \epsilon \ C_{m}$, we have that $ord(f(x))$ is a divisor of d.

I am currently repeating some algebra, however seem to be stuck how to solve this part of the exercise. I was thinking if I can show that $ord(f(x))$ divides both $n$ and $m$ it will also divide d. I just do not know how to get there. Any help would be greatly appreciated.

Answer 1

We know that $f(C_m)$ is a subgroup of $C_n$, so we have that $|f(C_m)| \big| n$. By first isomoprhism theorem we have that $|f(C_m)| \big| m$ and so $|f(C_m)| \big| \gcd(m,n) = d$.

So as the order of an element divides the order of the group we have that $\text{ord}(f(x))$ divides $|f(C_m)|$, which divides $d$