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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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    There is no number $a\ne0$ such that $a+a=0$. But there is a number $a\ne1$ such that $a\cdot a=1$. (I won't post this as an answer since (in effect) someone's already done that.)2011-09-26
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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 haven't learnt groups yet, is it possible to do this problem using just the definition of morphisms? and by the way what does R × /⟨−1⟩ mean? thanks2011-09-26
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    $\mathbb{R}^{\times}$ is the group of nonzero elements of $\mathbb{R}$ under multiplication and $\mathbb{R}^{\times}/ \langle -1 \rangle$ is a quotient group (don't sweat if you don't know what that means as it was an aside).2011-09-26
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    The hidden theorem used in my answer is that any isomorphism preserves order. Thus since $-1$ has finite order and all elements of $\mathbb{R}$ have infinite order there can be no isomorphism from $\mathbb{R}^{\times}$ to $\mathbb{R}.$2011-09-26
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    @Scharfschütze: I'm not sure what you mean when you say you know about morphisms but not about groups. You don't just have "morphisms": you have "morphisms in a certain category" or, more concretely, "morphisms preserving a certain structure". So when you say **isomorphic** you need to specify a certain structure. The two objects given *are* isomorphic as sets, i.e., there is a bijection between them. They are not isomorphic as groups, but if you don't know what a group is, how is it that you have been asked to show this??2011-09-26
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    @Pete L. Clark: I'm going to learn about groups this week but this is a homework problem I got last week(on semigroups and monoids). We haven't learnt isomorphism of groups yet nor have we looked at cyclic groups.2011-09-26
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    Groups is a full subcategory of Monoids. So if you want you can interpret isomorphism to mean module isomorphism. But the proof is the same.2011-09-26
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    and by module I meant moniod isomorphism2011-09-26
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    thanks!!! I think I understand you..so in R ×,<-1> is just {1,-1}, but in (R,+), -1 generates the entire group R, therefore they are not isopmorphic?2011-09-29
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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 $-1$.