Homomorphisms: Abstract Algebra?

Question

Given that m divides n, how do I find a homomorphism where phi(Z_n) = Z_m ?

Answer 1

As a minor abuse of notation, let a mod b denote the remainder of a when dividing by b. Then let φ:Z_n → Z_m be given by φ(a + nZ) = (a mod m) + mZ.

We must verify that this operation is well-defined -- suppose a + nZ = b + nZ, then a-b∈nZ, so n|a-b. Since m|n, we have m|a-b, so a mod m = b mod m, so (a mod m) + mZ = (b mod m) + mZ, thus φ is well-defined.

To see that it is a homomorphism, note that φ((a+nZ) + (b+nZ)) = φ((a+b) + nZ) = ((a+b) mod m) + mZ = (a mod m + b mod m) + mZ = ((a mod m) + mZ) + ((b mod m) + mZ) = φ(a+nZ) + φ(b+nZ), and likewise note that φ((a+nZ) * (b+nZ)) = φ(ab + nZ) = (ab mod m) + mZ = ((a mod m) * (b mod m)) + mZ = ((a mod m) + mZ) * ((b mod m) + mZ) = φ(a+nZ) * φ(b+nZ), so addition and multiplication are preserved.

Finally, note that any element of Z_m is of the form a+mZ for some a in the range 0≤a
Edit: sorry about that, I forgot that not everyone can read the unicode blackboard bold characters. I've replaced all of the blackboard bold Zs with ordinary Zs.

Answer 2

say that n = mp

phi(x) = x mod(m)
we have to show that phi((x+y)mod n) = phi(x) + phi(y)
say (x+y)mod n = z
x+y = kn + z.
therefore phi(x+y) = phi(kn+z) = phi(z) [since m divides n]
= z mod m.
phi(x) + phi(y) = x mod m + y mod m = (x+y)mod m = (kn+z)mod m = z mod m.