Let G = (Q - {0}, ), and let if be the subgroup H= {a/b | a and b are odd intege

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The fundamental theorem in the context of quotient groups says thefollowing: If : G-> H is a surjective group homomorphism andthe kernel of this homomorphism is K, then G/K is isomorphic toH. In this case, we wish to have G=Q*, H=Z, the integers underaddition, and K={a/b| a,b are both odd}. To start with, you must produce a suitable group homomorphism: Q*->Z. There are obviously many ways of getting one butthe point is to get one which is appropriate. Note that the kernelmust be only elements of the form a/b with ab being odd. Because of unique factorization of integers into prime powers, notethat for every non-zero rational a/b we can write a/b = ±(p1)n1(p2)n2.... (pk)nk where the n_i's are integers (positive or negative or zero)and the p_i 's are prime integers. Note that we may write thisproduct as an infinite one by putting power = 0 for the primes thatare not actually present, and this definition still makes sense.Thus for simplicity of notation, let us call for the prime p, the(unique) corresponding exponent np. Let (a/b)= n2 , the power corresponding to theprime 2. Note that such a map is clearly a homomorphism from Q* toZ, and is easily seen to be surjective, since (2i) = i for any integer i. Furthermore, which elements a/b would bein the kernel of ? Precisely those guys which haven2 = 0. But for such a/b 2 neither divides a, nor b(since we assume that a and b are relatively prime to start with)and that translates as saying a and b are both odd. This isprecisely H. Btw, it is a good exercise to see that H as describedis indeed a subgroup of Q*. This achieves what we sought.

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