Prove that every homomorphism f : Q → Z/mZ is trivial, i.e. f (x) = [0]m for all x ∈ Q.?
I've been working at this for hours, and can't even get a good start on it. I don't know whether to use the definition of a homomorphism to show that a non-trivial homomorphism wouldn't be well defined, or to show that every element would end up being mapped to 0...
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since f is a homomorphism, we have to have:
f(mx) = f(x + x +....+ x) (where x is added m times)
= mf(x) = (f(x) + f(x) +...+ f(x)) (m times) = 0
(since the order of any element of a subgroup of order m divides the order of the group).
but if x is in Q, say x = a/b, we can write x = ma/(mb) = m(a/(mb)).
thus f(x) = f(m(a/(mb)) = mf(a/(mb)) = 0, no matter what f(a/(mb)) may be.