Prove Ther Compute the kernel for the homomorphism : ZZ such that o(1) = 12. is {0}. Let ZZ and o(1) = 12. : Then ker() = {x Z\o(x) = 0}. Since (x)=(1+1+1+...x times) = x(1) and (1) = 12, then (x) = 12x. Then ker(){x Z/12x = 0}. Hence, we claim ker() = {0}. Also, by the homomorphism property, we can see that $(0) = 0. o(1) = 12, which we are given. (2) = (1 + 1) = (1) + (1) = 12 + 12 = 24. (3) = (1+1+1) = (1) + (1) + (1) = 12+12+12 = 36. etc. WTS ker() {0}. Let x ker(). Then (x) = 12x = 0 in Z. Let x=0k+r for some integer r such that 0