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

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    $\begingroup$ For any group $G$, if $G$ is generated by set $X$, $G=\langle X\rangle$, and $f,g\colon G\to K$ are two group homomorphisms, then $f=g$ if and only if $f(x)=g(x)$ for all $x\in X$. $\endgroup$ – Arturo Magidin Oct 21 '11 at 18:48
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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.

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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.

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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 $f([1])$.

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