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

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Note that if we're going to be considering this over more general algebraic structures, you need to ask yourself what $\sqrt\cdot$ means. – Jack M Aug 16 '14 at 0:17
If $a,b\in \mathbb Z$ then the operation is not even closed, and are you sure there are inverses? The identity is clearly $0$, so what is the inverse of, say, $1$? – Jack M Aug 16 '14 at 0:19
It's not possible for a commutative, associative operation with identity to have multiple inverses. Proof. Say that we have two inverses $x$ and $y$ for $a$. Then $x=x\circ(a\circ y)=(x\circ a)\circ y=y$, i.e. $x=y$. | By the way, you never stated what you think the identity of this operation is. – whacka Aug 16 '14 at 0:19
Our operation $\circ$ would be rather ugly and clunky if we had to define $\sqrt{}$ piecewise or with a branch cut or whatever in our domain. It doesn't make sense to pose the problem with $\sqrt{}$ left undefined; what is the domain in which it is most commonly canonically defined? That would be over the positive reals. So just take that to be your domain. Presumably this was an unstated assumption. – whacka Aug 16 '14 at 0:20
do you mean that there is an inverse if $a,b \in \mathbb{C}$ (instead of $a,b \in \mathbb{Z}$) ? – miracle173 Aug 16 '14 at 0:24

4 Answers

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 $\mathbb{C}$.
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Thanks for your sussinct answer Chris, can you expand a little more on what you mean by "Your set $\mathbb{H}$ together with the real numbers", I'm not sure what your'e stipulation together with the real numbers means ?. Thanks to evryon else for all of your responses. – Michael T Mckeon Aug 16 '14 at 1:50
@MichaelTMckeon The set you defined, $\mathbb{H} = \{x + iy \mid y > 0 ; x,y \in \mathbb{R} \}$, isn't a group under $\circ$. We have to add the identity element $0$, and the imaginary numbers need inverses. The latter can be either the positive or negative reals, and it's better to choose the positives, to be consistent with the usual definition of the square root symbol. Put it all together, and that's why I said "together with the nonnegative real numbers". – Chris Culter Aug 16 '14 at 1:55
I think I found a very simple answer to this problem which is to choose the field over which the function is defined to be the finite field modulo(2). So that the group becomes $(\circ, F_2)$, where $a \circ b = \sqrt{a^2 + b^2} mod(2)$ and $F_2=\{0,1\}$ – Michael T Mckeon Aug 18 '14 at 6:48

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 $0^2=0$.

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The trick is, can you give a non-trivial example of such an $R$? (I know of one example but it's a little esoteric!) – Steven Stadnicki Aug 16 '14 at 0:48
@StevenStadnicki Feel free to leave that example as either a comment, an answer, or even an edit to this answer. – Jack M Aug 16 '14 at 0:52
Jack: fair point! I'll add another answer here with the specific example I had in mind. (And I now realize I have no idea what this operation looks like there!) – Steven Stadnicki Aug 16 '14 at 0:55
Forgive my ignorance, but you are saying that this operator can be defined a a group for some kinds of rings $R$, but that the complex field is not suitable. For my proposed solution $a,b \in \mathbb{C}$ if $a = (x + iy)$ then $a^{-1} = (y - ix)$ or $a^{-1} = (-y + ix)$, how does this not satisfy the group axioms ?. – Michael T Mckeon Aug 16 '14 at 1:09
@MichaelTMckeon Actually, considered purely with the operation $\circ$, $\mathbb H$ is indeed a commutative group. – Jack M Aug 16 '14 at 1:51

To show an example of how Jack's construction works, here's a specific example:

It's well-known that the non-negative integers $\mathbb{Z}^{\geq 0}$ form a group under the operation $\oplus$ of nim-addition (or equivalently, under bitwise exclusive-or), with identity $0$ and each number being its own inverse. This infinite group (which is the projective limit of the product of $n$ copies of $\mathbb{Z}/2\mathbb{Z}$ as $n\to\infty$ in the obvious way) has finite subgroups corresponding to the restriction to $0\leq n\lt 2^i$ for each $i$.

What's less-known is that there's also a multiplication operation, nim-multiplication, which is compatible with nim-addition and forms a ring (in fact, a field!) over $\mathbb{Z}^{\geq 0}$. One way of defining the multiplication is to set $\alpha\otimes\beta = \mathop{mex}(\alpha'\otimes\beta\oplus\alpha\otimes\beta'\oplus\alpha'\otimes\beta' : \alpha'\lt\alpha, \beta'\lt\beta)$, where $\mathop{mex}()$ refers to the minimal excluded value, the smallest number not in the given set (under this notation, nim-addition itself can be defined by $\alpha\oplus\beta = \mathop{mex}(\alpha'\oplus\beta, \alpha\oplus\beta')$); another (admittedly more straightforward) way is as the projective limit of the Galois fields of order $2^{2^n}$; and much as in the addition case, the restrictions (in this case, the individual fields of size $2^{2^i}$) are all finite subfields. Here (swiped from Wikipedia) is the multiplication table for $n\lt 16 = 2^{2^2}$:

$$\begin{matrix} 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\ 0&2&3&1&8&10&11&9&12&14&15&13&4&6&7&5\\ 0&3&1&2&12&15&13&14&4&7&5&6&8&11&9&10\\ 0&4&8&12&6&2&14&10&11&15&3&7&13&9&5&1\\ 0&5&10&15&2&7&8&13&3&6&9&12&1&4&11&14\\ 0&6&11&13&14&8&5&3&7&1&12&10&9&15&2&4\\ 0&7&9&14&10&3&13&4&15&8&6&1&5&2&12&11\\ 0&8&12&4&11&3&6&15&13&5&1&9&6&14&10&2\\ 0&9&14&7&15&6&1&8&5&12&11&2&10&3&4&13\\ 0&10&15&5&3&9&12&6&1&11&14&4&2&8&13&7\\ 0&11&13&6&7&12&10&1&9&2&4&15&14&5&3&8\\ 0&12&4&8&13&1&9&5&6&10&2&14&11&7&15&3\\ 0&13&6&11&9&4&15&2&14&3&8&5&7&10&1&12\\ 0&14&7&9&5&11&2&12&10&4&13&3&15&1&8&6\\ 0&15&5&10&1&14&4&11&2&13&7&8&3&12&6&9\\ \end{matrix}$$ You can clearly see the subfield of order $4=2^{2^1}$ embedded in the top left corner of this multiplication table; you can also see that the diagonal is a permutation of $[0\ldots 15]$, or in other words that for each $x\in F_{16}$ there's a unique $y$ such that $y\otimes y=x$. (This isn't a proof for the arbitrary case, of course, but that can also be done; I believe Conway's On Numbers And Games covers the uniqueness of square roots). Determining what group of order 16 the operation $a\circ b = \sqrt[\otimes]{a^{\otimes2}\oplus b^{\otimes2}}$ generates is a nice exercise (hint: what's the inverse of $a$ under $\circ$?); in fact, $\circ$ here turns out to be even simpler than might be expected, thanks to a straightforward-but-unexpected identity! (another hint: expand $(a\oplus b)\otimes(a\oplus b)$.)

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Is this the same as the algebraic closure of $\mathbb Z_2$? – Jack M Aug 16 '14 at 1:30
@JackM Not quite, because there are polynomial equations that still have no solutions in the (containing) field - the Wikipedia article notes that $x^3+x+1$ has no roots. On the other hand, each successive field is a quadratic closure of the previous field, and so the infinite nim-field is quadratically closed; this is all covered wonderfully in ONAG. – Steven Stadnicki Aug 16 '14 at 1:34

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.)

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