Let G be the group of complex numbers with modulus 1 under multiplication? Is $\displaystyle G \cong G \times G$. Justify?
Do you understand how my hint gives you the result?...
Another approach would be to note that $\displaystyle G \lhd G \times G$ and so we can quotient out $\displaystyle G$ in both $\displaystyle G$ and $\displaystyle G \times G$.
I'm not sure where looking at the Kernel of the isomorphism would really get you. It made me conjour up the above paragraph, but I'm not sure what it would really do. However, in both of my solutions the actual group is irrelevent (as long as it is non-trivial), so I would suspect that there exists a solution that actually uses the group given...
Also, put the text [math-] and [/math-] (without the "-") around your maths to make it LeTeX-ey.
It’s not too hard to see that $\displaystyle G\times G\not\cong G$ as follows.
Suppose $\displaystyle \phi:G\times G\to G$ is a homomorphism. Then $\displaystyle \phi(1,1)=1$ since homomorphisms map identities to identities. Hence $\displaystyle 1=\phi(1,1)=\phi((-1,-1)(-1,-1))=(\phi(-1,-1))^2$ and so $\displaystyle \phi(-1,-1)=\pm1.$ Similarly $\displaystyle \phi(-1,1)=\phi(1,-1)=\pm1.$ It follows that $\displaystyle \phi$ maps the four-element set $\displaystyle \{(1,1),(-1,1),(1,-1),(-1,-1)\}$ to the two-element set $\displaystyle \{1,-1\}.$ So, what can you conclude here? Can $\displaystyle \phi$ be injective? Can it be an isomorphism?