Is there an extension of the complex plane that is a group under addition and multiplication? That is, having both $0$ and $1$ as the identity of $+$ and $\cdot$ respectively, and with an inverse for each ($-a$, $a^{-1}$) satisfying all the group laws.
EDIT:
Forming a group with $0$.
 A: The answer is no, in a very strong way. In particular, addition doesn't even enter the picture: the basic way that $0$ interacts multiplicatively with other numbers ruins any hope of producing a group in the way you want.
In $\mathbb{C}$ - or $\mathbb{R},\mathbb{Z},\mathbb{Q},...$ - $0$ has a special property: it is an annihilator for multiplication, that is, $x\cdot 0=0\cdot x=0$ for all $x$. This means in particular that in all these number systems, multiplication is not cancellative: $xz=yz$ does not in general imply $x=y$ (take $z=0$). (OK fine, this is specifically "right cancellability," but the point stands.)
The relevance of this is that in a group, the group operation is always cancellative. This is an immediate consequence of the existence of inverses and of associativity: if $xz=yz$ then $(xz)z^{-1}=(yz)z^{-1}$; but by associativity this means $x(zz^{-1})=y(zz^{-1})$, which in turn gives $x=y$. So no non-cancellable operation can be a group operation. 
Side note: cancellativity in fact means that for every group $(G,*)$ and every $g\in G$, the map $Mult_g:G\rightarrow G:h\mapsto gh$ is a permutation of $G$. In particular, every group can be thought of as a subgroup of the group of all permutations of some set, namely itself! This is Cayley's theorem, and also an important example of the concept of group actions (which if I recall correctly actually preceded the abstract notion of a group).
In particular, there's no way to extend $\mathbb{C}$ (or $\mathbb{R}$, or ...) to get a group under multiplication without changing the way multiplication works on $0$ even for "original" numbers.
