TITLE: Is $a \circ b = \sqrt{a^2+b^2}$ ever a group?
QUESTION [12 upvotes]: 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

REPLY [2 votes]: 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 $\mathbb{C}$.