I have an exercise to do and I don't understand how to solve it. It states:

Determine if the groups $(\mathbb{R} \!\,, +)$ and $(\mathbb{R} \!\,^*,\cdot{\,} \!\,)$ are isomorphic and to justify the answer.

I am not very good at this, but I would like a full demonstration.

$ \mathbb{R}^* = \mathbb{R}\backslash\{0\} $

Thank you very much!

  • 2
    $\begingroup$ Have you thought about torsion elements? $\endgroup$ – anon Nov 27 '13 at 22:21

$\textbf{Hint:} \:\:(-1)^2 = 1^2$

  • $\begingroup$ Thank you, but I would like to see an entire explanation. $\endgroup$ – Learner Nov 27 '13 at 21:55
  • 10
    $\begingroup$ You'll have to figure it out yourself. $\endgroup$ – Arthur Nov 27 '13 at 21:57
  • 9
    $\begingroup$ He gave you a property about one of the groups. Start by finding out which one he was talking about, and then check that this property would be conserved by a group isomorphism. Now, if you can prove this property doesn't hold in the other group, you've proven they are not isomorphic. $\endgroup$ – xavierm02 Nov 27 '13 at 22:13
  • $\begingroup$ I understood what you told me, but I would really appreciate a full answer in order to understand the full judgement. $\endgroup$ – Learner Nov 27 '13 at 22:16
  • 4
    $\begingroup$ If you understood what Arthur and xaviermo2 are saying then you should be able to do this problem. xaviermo2 has essentially outlined the entire solution. Please put some effort into solving the problem by yourself - you won't learn anything if they just spit out the entire solution for you. $\endgroup$ – Gyu Eun Lee Nov 27 '13 at 22:25

Regarding neat comment of @anon, note that: $$t(\mathbb R,+)=\{0\}\neq\{\pm1\}=t(\mathbb R^*,\cdot)$$ So you have different elements in both group which satisfy $x^n=id|_G$ where $n\in\mathbb N$.

  • $\begingroup$ Dear Babak...I'm facing the daily (anonymous) downvotes again :-( $\endgroup$ – amWhy Nov 30 '13 at 15:22

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