1. ## Isomorphism

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

2. 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!

3. ## order considerations

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