I have the following problem:
Let $C_m$, $C_n$ be two finite cyclic groups of cardinality $m$ and $n$ respectively. Let $d=\gcd(m,n)$ and consider the morphism of groups $$f:C_m\rightarrow C_n$$Show that for every $x\in C_m$, $ord(f(x))$ is a divisor of $d$
My Idea was the following:
Let us first fix two generators for $C_m,~C_n$ i.e.$$C_m=\langle g \rangle,~~~C_n=\langle h\rangle$$ Now $$f(x)^m=f(g^i)^m\stackrel{\text{f is a morphism}}{=}f(g^m)^i=f(1)^i=1~~~~~~~~~(1)$$ where $1\leq i\leq m-1$. Similarly since $f(x)\in C_n$ we know that $f(x)=h^j$ for some $1\leq j\leq n-1$. Therefore $$f(x)^n=(h^j)^n=1^j=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$$ But now from $(1)$ it follows that $ord(f(x))|m$ and similarly from $(2)$ that $ord(f(x))|n$ thus we are done.
Does this work so?
You can conclude from $(1)$ that the order of $f(x)$ divides $m.$
By LaGrange's Theorem and the fact that $f(x)\in C_n,$ you can conclude that the order of $f(x)$ also divides $n.$ (Do you see why?) At that point, you're basically done. Your work in $(2)$ almost does this, though I'd add $$...=\left(h^j\right)^n=h^{jn}=h^{nj}=\left(h^n\right)^j=1^j=....$$ The last equality above is where you use LaGrange's Theorem.