Let G be the group of complex numbers with modulus 1 under multiplication? Is $\displaystyle G \cong G \times G$. Justify?

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- Jun 7th 2009, 11:06 PMChandru1group theory question
Let G be the group of complex numbers with modulus 1 under multiplication? Is $\displaystyle G \cong G \times G$. Justify?

- Jun 7th 2009, 11:36 PMSwlabr
- Jun 7th 2009, 11:40 PMChandru1Answer
IF V is the vector space then

v \in V may be written as v=w1+w2 where W1 and W2 are subspaces and W1 \cap W2=(0).

Ok.

cant this problem be done by viewing something upon the Kernel. Like if it was isomorphic then K=e..so on - Jun 7th 2009, 11:58 PMSwlabr
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. - Jun 8th 2009, 01:09 AMTheAbstractionist
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?