Undergrad Showing that two groups are not isomorphic question

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To demonstrate that the groups ##\mathbb{R} - \{0\}## and ##\mathbb{C} - \{0\}## are not isomorphic, the discussion highlights the differing number of solutions to the equation ##x^3 = 1##, with one solution in the reals and three in the complex numbers. The use of the equation ##x^2 = -1## is suggested as a less straightforward method, raising concerns about preserving properties under isomorphism. The finite order elements are also considered, noting that ##\mathbb{R} - \{0\}## has only two while ##\mathbb{C} - \{0\}## has infinitely many. The conversation touches on the complexities of assuming an isomorphism, particularly regarding order preservation. Ultimately, the discussion leads to a potential resolution involving the product of elements, specifically considering ##\varphi(-1) \cdot \varphi(-1)##.
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I am trying to show that ##\mathbb{R} - \{ 0\}## is not isomorphic to ##\mathbb{C} - \{0 \}##. If we simply look at ##x^3 = 1##, it's clear that ##\mathbb{R} - \{ 0\}## has one solution while ##\mathbb{C} - \{0 \}## has three.

My question, how can I use ##x^2 = -1## to show that they are not isomorphic? Using ##x^3 = 1## is more clear because any isomorphism would preserve powers and preserve the identity. But using ##x^2 = -1## is less clear to me.
 
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Interesting question. I would consider all elements of finite order. There are only two of them in ##\mathbb{R}-\{0\}## and infinitely many in ##\mathbb{C}-\{0\}##. But if you only want to use the single relation ##\varphi(r)=i## it's a bit tricky, because one easily falls into unproven statements like the ordering of the two sets and an assumed isomorphism isn't necessarily order preserving. Do you have any ideas?

Edit: I think I got it: Consider ##\varphi(-1)\cdot \varphi (-1)##.
 
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