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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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    @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).2016-07-05

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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.

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    @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.2011-09-01
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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?

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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.2011-09-01
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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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    @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.2011-09-02
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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.