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Am I incorrect in believing that the following exercise is not possible?

Prove that the additive group $\mathbb{R}^+$ of real numbers is isomorphic to the multiplicative group $P$ of positive reals.

My reasoning is that if we had $\phi \colon \mathbb{R}^+ \rightarrow P$ as our isomorphism, then we have

$$\phi(\frac{1}{3}) \longmapsto \frac{1}{3}$$ and $$\phi(-3) \longmapsto \frac{1}{3}$$

Am I missing something?

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Why would we have that? –  Qiaochu Yuan Dec 15 '12 at 9:51
@QiaochuYuan Do we not have to map inverses to inverses? That along with the fact that $\frac{1}{3} \in \mathbb{R}^+$ and $P$. –  providence Dec 15 '12 at 10:00
Both of those things are true, but they don't imply what you're concluding. If you write out your argument in more detail it should be easier to see the mistake. –  Qiaochu Yuan Dec 15 '12 at 10:05
@QiaochuYuan Oh, right, reals. I've just come off a binge of problems dealing with $\mathbb{Z}$ and $\mathbb{Q}$ exclusively. I see my error. –  providence Dec 15 '12 at 10:14

1 Answer 1

up vote 3 down vote accepted

Hint: How do $e^{x+y}$ and $e^x \cdot e^y$ relate?

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Isn't that an isomorphism between $\mathbb{R}^+$ and the cyclic subgroup of $\mathbb{R}$ generated by $e$? –  providence Dec 15 '12 at 9:58
@providence: nope. –  Qiaochu Yuan Dec 15 '12 at 10:04
@providence: Note that $(\mathbb R,+)$ and $(\mathbb R^*,.)$ are not the same. –  Babak S. Dec 15 '12 at 10:10

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