Isomorphism

**Jimmy_W** · April 20th 2009, 01:27 PM

Prove that $(\mathbb{R} \$ \{0} $) * (\mathbb{R}$ \{0}) is not isomorphic to $\mathbb{C}$ \{0} (the group operation in both cases is multiplication).

**NonCommAlg** · April 20th 2009, 01:44 PM

Originally Posted by Jimmy_W

Prove that $(\mathbb{R} \$ \{0} $) * (\mathbb{R}$ \{0}) is not isomorphic to $\mathbb{C}$ \{0} (the group operation in both cases is multiplication).

suppose there was an isomorphism $f.$ let $f(x,y)=i$ and $f(a,b)=-1.$ then $f(a^2,b^2)=1=f(1,1).$ thus $a^2=b^2=1.$ hence at least one of $a,b$ is equal to $-1.$ why?

but then $f(x^2,y^2)=i^2=-1=f(a,b)$ and so $x^2=a, \ y^2=b.$ hence at least one of $x^2,y^2$ is equal to $-1,$ which is obviously impossible!

**Gamma** · April 20th 2009, 01:51 PM

You can just say these groups cannot be isomorphic due to the fact that $\mathbb{C}^{\times}$ has elements of order 4, namely $i, -i$ while there are no elements of order 4 in $\mathbb{R}^{\times}\times \mathbb{R}^{\times}$

everything in $\mathbb{R}^{\times}\times \mathbb{R}^{\times}$ has order
one: (1,1)
two: (-1,1), (1,-1) and (-1,-1)
or
infinite order: everything else

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