I have the following problem:
Let $C_m$ and $C_n$ be two finite cyclic groups of cardinality $m$ and $n$. Let $d=\gcd(m,n)$ and $f:C_m\rightarrow C_n$ be a morphism of groups. Let $C_d\subset C_n$ be the unique subgroup of cardinality $d$. Show that the image of $f$ is contained in $C_d$.
I have just shown in a first part that $\forall x\in C_m$, $ord(f(x))|d$.
And I wanted to prove the statement as follows:
Proof
Let $g$ be a generator of $C_n$, i.e. $$\langle g \rangle=\{1,g^1,g^2,...,g^{n-1}\}=C_n$$Then since $C_d$ is a subgroup of $C_n$ we know that also $C_d$ is cyclic, to be more precise $$\{1,g^1,...,g^{d-1}\}=C_d$$ where $d<n$. Now let $x\in C_m$ and consider $f(x)$. Then since $f$ is a morphism of groups $$Im(f)<C_n\Rightarrow f(x)\in C_n \Rightarrow f(x)=g^i$$ for some $1\leq i\leq n-1$, but we just know that $ord(f(x))|d$ so we indeed have that $f(x)=g^j, \,\,1\leq j\leq d-1$. And this means that $f(x)\in C_d$, and thus $Im(f)\subset C_d$.
Does this work or is it completly wrong? It would be helpful if someone could take a look and write some comments.
Thanks a lot.
$Im(f)$ is a subgroup of $C_n$. So all elements of $Im(f)$ have order dividing $n$.
Let $x\in C_m$ with order $l$. Then $l$ divides $m$. Thus $f(x)^l = f(x^l) = f(1) = 1$ and so $f(x)$ has order dividing $l$ which divides $m$.
It follows that all elements of $f(C_m)$ have order dividing both $m$ and $n$ and so have order dividing $\gcd(m,n)$.