Date: 02/11/2003 at 19:16:37 From: Scott Subject: Isomorphic groups Is the additive group of rationals isomorphic to the multiplicative group of non-zero rationals? TO prove isomorphism I must find a bijective map such that f(xy) = f (x)f(y). However, I can't find one that will map reals to real and preserve that 0 must get mapped to 1. I think that they are not isomorphic but cannot locate anything in their structure that is different.
Date: 02/12/2003 at 02:56:37 From: Doctor Jacques Subject: Re: Isomorphic groups Hi Scott, Assume that there is such an isomorphism. Then there is a rational q such that f(q) = 2. Now, q/2 is also rational, and we should have: f(q/2 + q/2) = f(q/2)*f(q/2) = 2 which implies that f(q/2) = sqrt (2) or -sqrt(2). As neither of these is rational, we reach a contradiction - there is no such isomorphism. Note: I could also have used -1 instead of 2, but this proves that the additive group is not even isomorphic to the multiplicative group of non-zero _positive_ rationals. If you replace "rationals" by "reals" in this last case, there is such an isomorphism, namely the exponential function. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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