Conditions for property of function $\Bbb Z/m\Bbb Z \to \Bbb Z/n\Bbb Z$

Question

I've added a red star next to the line I don't understand, I also don't understand where the condition came from.

Answer 1

Remember that in $\mathbb{Z}/n\mathbb{Z}$, $\underbrace{1+_n\cdots +_n1}_{n}=0$. If you want $\varphi(k+_m l)=\varphi(k)+_n\varphi(l)$ (because you want a group homomorphism), in particular you need it for $k=l=1$. The reminder of devision of $1$ by any number greater than $1$ is again $1$, so that is where the equation

$$0=\varphi(0)=\varphi(\underbrace{1+_n\cdots +_n1}_{n})=\underbrace{1+_m\cdots +_m1}_{n}$$

comes frome. Adding $1$'s in $\mathbb{Z}/n\mathbb{Z}$ only sum up to $0$ if the times you add $1$ is a multiple of $n$. In this case, $m$ must be a multiple of $n$, which is the same as saying that $n$ must divide $m$.

Next, I'll show that this condition is sufficient. You can think of elements of $\mathbb{Z}/m\mathbb{Z}$ as equivalence classes of integers with the same reminder on the division by $m$, so take $\overline{k}=k+rm$ and $\overline{l}=l+sm$. Since $m=hn$, we have $\overline{k}=k+rhn$ and $\overline{l}=l+shn$. The homomorphism $\varphi$ is just sending each element of $\mathbb{Z}/m\mathbb{Z}$ to itself modulo $n$. Hence, denoting equivalence classes by $[]_m$,

$\varphi([\overline{k}]_m+_m[\overline{l}]_m)=\varphi([\overline{k}+\overline{l}]_m)=[\overline{k}+\overline{l}]_n=[k+rhn+l+shn]_n=[k+l]_n=k+_n l$

I've used that the sum is not dependent on the representative of the equivalence, and the clear fact that taking the reminder of the division by $n$ doesn't depend on the representative.

Answer 2

The remainder when you divide $0$ by $n$ is $0$ so $\varphi(0) = 0$. You know that $m \equiv 0 \mod m$ so $\varphi(m)=0$. Now you just write $m$ as $1+_m\ldots+_m 1$ ($m$ times) so you have $\varphi(1+_m\ldots+_m 1) = 0$. Assuming the homomorphism property you get $\varphi(1) +_n \ldots +_n \varphi(1)=0$. Since dividing 1 by $n$ you always get a remainder of $1$ as long as $n>1$, $\varphi(1) = 1$. That means that $1 +_n \ldots +_n 1 = 0$. This sum is in the group $\mathbb{Z}/n\mathbb{Z}$ but we still have $m$ summands so this says that the $m \equiv 0 \mod n$ which means $m$ must be a multiple of $n$.