We have the Cartesian product $B \times B$ and there we have the addition $$(f,g)+(h,k)=(f+h,g+k)$$ and the multiplication $$(f,g) \cdot (h,k)=(f\cdot h+g\cdot k,f\cdot k+g\cdot h).$$ I want to find the identity element of the group related to the addition. So I must have something like $(f,g)+e=(f,g)$. What do I do now? Is it something like ($f+e,g)=(f,g)$?

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Of course $e$ is also a pair, $e=(e_1,e_2)$. What can you conclude from $(f,g)+(e_1,e_2)=(f,g)$?
Assuming $B$ is a group, it has an identity element $e$. In $B \times B$, therefore $$(f,g) + (e,e) = (f + e, g + e) = (f,g)$$ for any elements $f$ and $g$, so $(e,e)$ is the identity in $B \times B$.
You don't determine what $e$ is. $e$ is supposed to be the identity element of $B$ with respect to addition. By this calculation $(e,e)$ is the identity in $B \times B$. – Hans Giebenrath Dec 1 '12 at 19:11
@Youmath If by "$0$" you mean "the identity element of $B$ with respect to addition", then yes. You want to be careful with that terminology, however. The group $B$ could be strange and not have an element called "$0$". It might not have any numbers at all. – Austin Mohr Dec 1 '12 at 19:27