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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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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 $f(x*y) = f(x) + f(y)$. –  PseudoNeo Sep 1 '11 at 17:35
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God forbid you get inappropriate answers... –  The Chaz 2.0 Sep 1 '11 at 17:39
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Given an operation, how can I determine if it involves addition or multiplication? –  Charles Sep 1 '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 1 '11 at 18:03
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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 1 '11 at 19:06

5 Answers 5

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.

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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 1 '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 1 '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 1 '11 at 19:24
    
@Bill Dubuque: Done. I think. –  André Nicolas Sep 1 '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 1 '11 at 19:37

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?

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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 1 '11 at 19:50

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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What is the binary operation in this Klein group? –  Primeczar Sep 1 '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 1 '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 1 '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 en.wikipedia.org/wiki/Klein_four-group –  Ross Millikan Sep 1 '11 at 18:08
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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 2 '11 at 0:56

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.

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Relatively close to Ross Millikan's suggestion, but given in different language: The natural numbers form a group under bitwise XOR. I think that (using 2s complement representation for negative integers) you can actually include all the integers.

But really. Tell us what's wrong with the suggestions?

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OP asked for all reals, not a countable set. –  Ross Millikan Sep 1 '11 at 18:44

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