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I think it wants me to show that there's no bijection between the two sets.

I first tried to show that there's simply no bijection, then I realised that it doesn't work.

If I'm to show there's no bijective morphism that carries multiplication in nonzero real numbers to addition in real numbers, how am I supposed to do it? I'm thinking about doing something with 0, since it's the element of the first set but not the second set.

Thanks!

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    I thought of that in the beginning, but I don't think that works because you can always use f to assign 0 to any element.2011-09-26

2 Answers 2

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In $(\mathbb{R}, +)$ there are no torsion elements (the group generated by any non-identity element is infinite cyclic). On the other hand, in $\mathbb{R}^{\times},$ the group $\langle -1 \rangle$ is finite.

That said, it is interesting to note $\mathbb{R}^{\times}/ \langle -1 \rangle \cong (\mathbb{R}_{+}, \cdot)$ which is isomorphic to $(\mathbb{R}, +)$ via the natural logarithm. In fact, $\mathbb{R}^{\times} \cong \mathbb{Z}/2 \oplus \mathbb{R}.$

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    I actually thought of a proof that doesn't involve cyclic groups. let f be an isomorphism, then f(-1*-1*-1)=f(-1), and f(-1*-1*-1)=f(-1)+f(-1)+f(-1)=3f(-1), so 3f(-1)=f(-1), so f(-1)=0. Since morphism of groups is also a morphism of units, f(1)=0. But then both f(-1) and f(1)=0, so f is not a bijection. Contradiction. So there's no isomorphism between these two groups. Is my proof OK?2011-09-29
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Hint: there is something special about $-1$.