Is $a \circ b = \sqrt{a^2+b^2}$ ever a group? I've been asked to "Show that the operation $a ~ \circ ~ b = \sqrt{a^2+b^2}$ is associative, is commutative and has an identity but that the inverses do not always exist."
Wihch is easy enough to do if it is assumed that  $a,b \in \mathbb{R}$, except that the question does not stipulate over what field a and b exist, If however $a,b \in \mathbb{C}$ then all elements do have invereses in fact each element has two inverses, which I don't think is appropiate for a group, but this started me thinking maybe this operator might be classifable as a group operation if $a,b$ are chosen to exists over some specific field such as matrix, modular or finite or other field or ring ?. 
One possibility I have considered is complex upper half plane defined by $a,b \in \mathbb{H} = \{x + iy \mid y > 0 ; x,y \in \mathbb{R} \}$  but I am not sure if this would suffice to define the operation as a group operator or to prove the question incorrect. 
I believe the question is meant to be contradictory and stimulate discussion, but I am not knowledgable enough to determine if there are any situations in which this operator will suffice for a group ?, any help would be greatly appreciated.
Michael
 A: Let $R$ be a ring* such that for each $x\in R$ there is a unique element $x'$ such that $x'^2= x$, call that element $\sqrt x$.
Then the operation $a\circ b=\sqrt{a^2+b^2}$ can be defined, is associative and commutative, and has the identity $0$ (the additive identity of $R$). Now suppose for an element $a$ there is a $b$ such that $a\circ b=0$, that is, $a$ has an inverse. It follows from the definition of $\sqrt\cdot$ and the ring axioms that $\sqrt 0=0$, so we must have $a^2+b^2=0$ and $b^2=-a^2$. Thus $a$ has an inverse for $\circ$ iff $-a^2$ is the square of some element, but this is guaranteed by our assumptions about $R$.
Thus, as long as $\sqrt\cdot$ is uniquely defined and $R$ is closed under it, $R$ is a group under $\circ$. Note that uniqueness really is important, since if $\sqrt\cdot$ isn't unique, there isn't even an identity. If $x\neq y$ both have the same square, then either $x\circ0\neq x$ or $y\circ0\neq y$. This eliminates the possibility of just taking something like $\mathbb C$ and defining $\sqrt\cdot$ piecewise.
* Technically, we don't need distributivity. $R$ needs to be a group under addition and satisfy $0^2=0$.
A: To address the second paragraph of the question: Yes, $\circ$ is a group operation on any of the following subsets of $\mathbb{C}$:


*

*$\{0\}$

*Your set $\mathbb{H}$, together with the nonnegative real numbers

*More generally, $\sqrt{G}$ where $\sqrt{\phantom{G}}$ is a branch of the square root and $G$ is a subgroup of $\mathbb{C}$.

A: I don't know if this is what you're looking for, but if you are willing to accept extending the monoid in question, the Grothendieck group of this monoid can be constructed. I haven't worked out the details, but I believe the result could be quite interesting.
(You might want to restrict the monoid you start with to be the non-negative integers/rationals/real numbers if you want the cancellation property.)
