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In my last question I asked for examples of groups formed by real numbers where the operation is something different from addition or multiplication. With these words I think I could not convey what I wanted. In an attempt for further clarity in conveying my query I state the question as follow

" Are there examples of groups formed by real numbers where the binary operation of the group does not involve any addition or multiplication" I hope this time I will be getting appropriate answers.

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Primeczar
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  • Do you need the elements of the group to be all real numbers, or just a subset? – Charles Sep 01 '11 at 17:29
  • Also, what does "involves" mean? – Charles Sep 01 '11 at 17:31
  • @charles - Yes I want groups with the mentioned property whose elements are all real numbers. – Primeczar Sep 01 '11 at 17:35
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    It's not an answer, but one can show that any $C^1$ group structure on $\mathbb R$ (that is, a group structure whose multiplication $$ and inverse are continuously differentiable functions) is in fact the addition in disguise (precisely: there is a $C^1$-diffeomorphism $f$ of $\mathbb R$ such that $f(xy) = f(x) + f(y)$. – PseudoNeo Sep 01 '11 at 17:35
  • @charles- I meant the binary operation should not be multiplication. – Primeczar Sep 01 '11 at 17:36
  • For example, a+b, or a.b or a+b+(a.b), in a nutshell there should not be any addition or multiplication . – Primeczar Sep 01 '11 at 17:37
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    God forbid you get inappropriate answers... – The Chaz 2.0 Sep 01 '11 at 17:39
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    Given an operation, how can I determine if it involves addition or multiplication? – Charles Sep 01 '11 at 17:41
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    Just to insist on The Chaz's irony, I find your opening of three questions on MSe/MO and the comments “I hope this time I will be getting appropriate answers” (here) and ”I hope in this site I may get better answers as compared to its stack exchange counterpart” (in MO) insulting. – PseudoNeo Sep 01 '11 at 18:03
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    HINT $\ $ Take any group $:G:$ with the same cardinality as $:\mathbb R:$ and transport the group structure of $:G:$ to $:\mathbb R:$ along any bijection of their underlying sets. – Bill Dubuque Sep 01 '11 at 18:15
  • You do know that $\mathbb{R}$ is by itself just a continuum, right? – Aleksei Averchenko Sep 01 '11 at 18:34
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    Rather bad form to open a second question with identical title and which is nothing but an attempt at explaining what you meant with the original. Much better would have been to edit your old question to add your clarification. As it is, I have voted to close the old question as a duplicate, and invite you to not do this again in the future. – Arturo Magidin Sep 01 '11 at 19:06
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    @PseudoNeo do you have a source for your statement about $C_1$ group operations on $\Bbb R$? – Mario Carneiro Jun 20 '16 at 08:22
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    @MarioCarneiro: Well, I've learnt about it in a French exercise book on differential calculus (Rouvière, Petit guide de calcul différentiel). After your question I googled a bit and found this Conrad blurb: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/relativity.pdf which may suit you. (These blurbs are so amazing that one of these days I will launch a kickstarter project to make a bronze statue of K. Conrad). – PseudoNeo Jul 05 '16 at 09:59

4 Answers4

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Let $W$ be the collection of all bijections from the natural numbers $\mathbb{N}$ to $\mathbb{N}$. It is a standard fact that the cardinality of $W$ is the same as the cardinality of $\mathbb{R}$.

It follows that there is a bijection $\phi: \mathbb{R}\to W$. Instead of using the notation $\phi(a)$, we will use the perhaps clearer notation $\phi_a$

For any real numbers $a$, $b$, define $a\ast b$ as follows. $$a\ast b=\phi^{-1}(\phi_a\circ \phi_b).$$

Note that $\phi_a$ and $\phi_b$ are bijections from $\mathbb{N}$ to $\mathbb{N}$, and $\phi_a\circ\phi_b$ is the composition of the functions $\phi_a$ and $\phi_b$, defined by $$(\phi_a\circ\phi_b)(n)=\phi_a(\phi_b(n)),$$ (apply $\phi_b$, then apply $\phi_a$ to the result). It is clear that $\phi_a\circ\phi_b$ is a bijection.

It is not hard to verify that under the operation $\ast$, the real numbers form a group, indeed a very non-abelian group. Ordinary sum and product are nowhere involved in the definition of $\ast$.

Comment: The above answer is a special case of the general construction method "You can realize any group whose cardinality is the continuum this way" in the answer of Yuval Filmus.

André Nicolas
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  • That's simply applying transport of structure to a specific group of cardinality $\mathbb R$ - which has already been mentioned by Yuval. Of course one can do exactly the same for any group of cardinality $\mathbb R$, as Yuval mentioned. – Bill Dubuque Sep 01 '11 at 19:04
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    Agreed! However, the OP seems to want to avoid any mention of conventional addition and multiplication. The example was designed to do precisely that. "Scrambling" a group of cardinality $c$ that is based on addition and/or multiplication might not qualify as an example to the OP. – André Nicolas Sep 01 '11 at 19:15
  • While I agree that Yuval's first paragraph doesn't meet the goal of "not involving addition", I interpret his second paragraph as referring to general transport of structure. It would be helpful to mention that terminology, and to stress that this answer is a specific case of this general technique alluded to in Yuval's answer. Else the relationship may not be clear to students. Such matters are often confusing to students attempting to first grasp the concepts of algebraic structure and isomorphism. – Bill Dubuque Sep 01 '11 at 19:24
  • @Bill Dubuque: Done. I think. – André Nicolas Sep 01 '11 at 19:31
  • Thanks. Of course I think it is a good idea to present a specific example as you do. My only concern is that students may not realize that both answers are examples of the general technique of transporting algebraic structure along a bijection (set-theoretic isomorphism) - one that is rather trivial from an algebraic perspective. Alas, this takes some nontrivial algebraic experience to appreciate, so it is difficult to convey to beginning students. – Bill Dubuque Sep 01 '11 at 19:37
  • @Bill, André: Is this really a special case of Yuval's construction? Unless I'm missing something, the latter always yields an abelian group, while this one does not. –  Sep 01 '11 at 19:40
  • @Rahul As I said, I read Yuval's second paragraph as saying that one can transport the group structure of any group of same cardinality to the set $\mathbb R$. That's precisely what Andre does - for one specific group. – Bill Dubuque Sep 01 '11 at 19:47
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Choose a permutation $\pi$ of the real numbers, and define a group using $f(a,b) = \pi^{-1}(\pi(a) + \pi(b))$. While this "involves" addition, if you don't know $\pi$, then the operation would look quite random.

You can realize any group whose cardinality is the continuum this way. Which of them would you consider addition-like or multiplication-like?

Yuval Filmus
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    It would help to stress in your second paragraph that transporting the group structure from some other group to the set $\mathbb R$ does not "involve" the addition of $\mathbb R$ but, rather, the group operation of the other group. The example in your first paragraph may mislead readers to think otherwise. – Bill Dubuque Sep 01 '11 at 19:50
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Please link to the other question. What was wrong with the idea to take your favorite group and name each element in the operation table with a real number? For example, the Klein group: $$\begin {array} {cccc} 0&\sqrt{2}&5.3&\pi \\\sqrt{2}&0&\pi&5.3\\5.3&\pi&0&\sqrt{2}\\ \pi&5.3&\sqrt{2}&0 \end{array}$$ No addition or multiplication in sight. Silly, perhaps, but I don't understand what you are looking for, so maybe this will help define the question.

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Ross Millikan
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  • What is the binary operation in this Klein group? – Primeczar Sep 01 '11 at 17:42
  • @Ross: This would be a good answer, but Primeczar wants a group operating on R, not a subset of R. – Charles Sep 01 '11 at 17:45
  • I saw "composed of real numbers" as "some real numbers", but I can see he said after I posted that he wants "all real numbers". – Ross Millikan Sep 01 '11 at 17:51
  • @Primeczar: The table gives the operation. For a finite table you don't need a rule, you can just give the result. I named the elements with various real numbers. You can see it at http://en.wikipedia.org/wiki/Klein_four-group – Ross Millikan Sep 01 '11 at 18:08
  • I think, on reflection, that this deserves a +1 despite not meeting the comment-requirement to use all of R. – Charles Sep 01 '11 at 19:39
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    @Primeczar As Ross basically indicates, in some sense, the table and the binary operation, which I'll denote "G", don't differ at all. You might say the binary operation is the table even. You can also view it as a rule if you wish, which I outline as follows: {If x=0, and y=0, then (xGy)=0. If x=0, and y=5.3, then (xGy)=5.3...} In other words, as a rule, you can interpret the binary operation as a rule which consists of a collection of "If x=a, and y=b, then (xGy)=z" rules. – Doug Spoonwood Sep 02 '11 at 00:56
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How about identify the reals in $(0,1)$ with the canonical set of subsets of $\mathbb{N}$? Of course you need to deal with the ambiguity of trailing 0 versions and trailing 1 versions of the terminating ones-a bijection will solve that. Then use symmetric set difference operation as your operation. Biject $(0,1)$ with $\mathbb{R}$ in your favorite way-mine is arctangents.

Ross Millikan
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