from Z12 to Z28 , How many homomorphism is possible.
how it can be done?
Z12 is cyclic, so any homomorphism is completely determined by the image of an generator:
(if G = <x>, so that every g in G is equal to xk for some k, then for any homomorphism h:G→G', we have:
h(xk) = (h(x))k).
in particular, any homomorphism h from Z12→Z28 is completely determined by h(1).
things to prove first:
the order of h(x) in Z28 must divide the order of x in Z12.
this means (once you have PROVED this) that |h(1)| must be a divisor of 12: 1,2,3,4,6 or 12.
but h(1) is also an element of Z28, so its order must also divide 28.
3,6 and 12 do not divide 28, which means that h(1) must have order 1,2 or 4.
prove if h(1) has order 1, then h(1) = 0, and thus h(x) = 0 for all x in Z12.
next, prove that if h(1) has order 2, then h(1) = 14, and thus h(x) = 14x (mod 28). how big is the image of h?
finally, prove that if h(1) has order 4, then h(1) = 7, or h(1) = 21. show that this gives two different homomorphisms into the same subgroup of Z28.
how many homomorphisms is that?
the answer isn't a "formula". the answer is based on "understanding". i'll repeat my self for emphasis:
WHY is it true that h(xk) = (h(x))k, for a homomorphism h?
WHY does that mean we only need to find h(1)?
WHY does the order of h(x) divide the order of x?