Prove that $\displaystyle (\mathbb{R} \ $\{0}$\displaystyle ) * (\mathbb{R}$\{0}) is not isomorphic to $\displaystyle \mathbb{C}$\{0} (the group operation in both cases is multiplication).
suppose there was an isomorphism $\displaystyle f.$ let $\displaystyle f(x,y)=i$ and $\displaystyle f(a,b)=-1.$ then $\displaystyle f(a^2,b^2)=1=f(1,1).$ thus $\displaystyle a^2=b^2=1.$ hence at least one of $\displaystyle a,b$ is equal to $\displaystyle -1.$ why?
but then $\displaystyle f(x^2,y^2)=i^2=-1=f(a,b)$ and so $\displaystyle x^2=a, \ y^2=b.$ hence at least one of $\displaystyle x^2,y^2$ is equal to $\displaystyle -1,$ which is obviously impossible!
You can just say these groups cannot be isomorphic due to the fact that $\displaystyle \mathbb{C}^{\times}$ has elements of order 4, namely $\displaystyle i, -i$ while there are no elements of order 4 in $\displaystyle \mathbb{R}^{\times}\times \mathbb{R}^{\times}$
everything in $\displaystyle \mathbb{R}^{\times}\times \mathbb{R}^{\times}$ has order
one: (1,1)
two: (-1,1), (1,-1) and (-1,-1)
or
infinite order: everything else