Let m and n be positive integers with n dividing m. Prove that the natural surjective ring projection Z/mZ ->Z/nZ is also surjective on the units:(Z/mZ)^x ->(Z/nZ)^x
Since n divides m, then m=kn for some positive integer k. The natural surjective ring project
f: Z/mZ -> Z/nZ is defined as f(a)=(a mod n) for each a in Z/mZ.
Now we need to show that f is also surjective between units of Z/mz and Z/nZ
We know, a is a unit of ...
Surjective Ring Projections are investigated.