Homomorphisms between $\Bbb Z_n$ and $\Bbb Z_m$ [closed]

Solution 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.

Solution 2:

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 $f([1])$.