Homomorphisms between $\Bbb Z_n$ and $\Bbb Z_m$

Question

Let's look at $f:\Bbb Z_{10}\to \Bbb Z_{12}$ . How do you prove that $f$ is uniquely determined by what $f[1]$ is? What is the order of $f[1]$?

Answer 1

Hint: Because $f$ is a homomorphism, $$f(n+m)=f(n)+f(m)$$ Use this to determine $f(n)$ in terms of $f(1)$, for any $n\in Z_{10}$.

Also, recall the definition of the order of an element of a group. The order of $f(1)$ is just its order as an element of $Z_{12}$. Consider: can we choose any element of $Z_{12}$ to be $f(1)$? Note that $$\underbrace{f(1)+\cdots+f(1)}_{10\text{ times}}=f(10)=f(0)=0\in Z_{12}$$ because $10=0$ in $Z_{10}$ and $f(0)=0$ because $f$ is a homomorpism.

Answer 2

Since each of these groups are cyclic, with generator $[1]$, it will determine the rest. So, if $[m]\in\mathbb{Z}/10$ we have that $[m]=[1]+[1]+\cdots +[1]$ where $[1]$ is added $m$ times. Then

$$ f([m])=f([1]+\cdots +[1])=f([1])+\cdots +f([1]), $$

since $f$ is a homomorphism. But we know what $f([1])$ is and so $f([m])$ is determined. The order of $f([1])$ is the order of whatever element $[1]$ gets mapped to.

Answer 3

In addition the other answers, note that both $n$ and $m$ constrain the order of $f([1])$ because $0 = f([0]) = f(n \cdot [1]) = n \cdot f([1])$. Since we also have $0 = m \cdot f([1])$, we get $0 = \gcd(m,n) \cdot f([1])$. This gives an upper bound to the order of $f([1])$.