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Considering a suitable set of numbers, you can construct a group using addition, and you can construct another group using multiplication. My question: Can you construct a group using exponentiation?

  • $(\mathbb{R}, +)$ is a group.

  • $(\mathbb{R} \backslash 0, \times)$ is another group.

  • Does there exist some $S \subseteq \mathbb{R}$ such that $(S, \uparrow)$ is a group?

(Here $x \uparrow y = x^y$.)

I'm going to guess "no", since exponentiation is asymmetric and hence needs two inverses (roots and logarithms). So maybe you can have some other group-like structure? (But which one?)

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    You can have nonabelian groups. The main problem is that $\uparrow$ is not associative.2012-05-11
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    Oh man, I hadn't even _thought_ about associativity!2012-05-11
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    ;) You cannot go very far without it.2012-05-11
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    what happens if you consider ${0,1}$ as the candidates .2012-05-11
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    oh it doesn't help :P2012-05-11
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    Whether a group's binary operation is denoted $+$ or $\cdot$ is a notational convention that fits the better (more intuitive) frame of mind for a particular context. If we really wanted to, we could denote addition by $\cdot$ or multiplication by $+$, perhaps in a parallel universe, and there would be no issue. The real question is, in what conceptual sense could an arbitrary associative binary operation on a set be thought of as "exponentiation"? Fundamentally, I think exponentiation's moral purpose is to merely transfer from one structure to another - not build one from scratch.2012-05-11

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$x^y$ is not even well-defined for many real arguments. (For example, $(-2)^\frac12$.) You would have to begin by restricting the domain to non-negative reals.

But even then, exponentiation is not in general associative, since $x^{(y^z)} \ne (x^y)^z$, and to make it associative you would have to restrict the domain so severely that what is left would not be very interesting.

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    Not associative. Definitely not a group then. Perhaps some kind of loop or quasigroup?2012-05-11
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    Well, if you restrict it to the set $\{1\}$2012-05-11
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This isn't quite what you want, but it's a nice example and deserves to be widely known: the reals exceeding 1 form a (commutative!) group under the operation given by $a*b=a^{\log b}$.

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    Is that $>1$ or $\geq 1$? It seems like $1$ would serve as the identity, but you wrote reals exceeding one.2012-05-12
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    @chris the base of the logarithm. If you chose $\ln$ it will be $e$.2012-05-12
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    I realized after a bit more thought my earlier comment was incorrect. Thanks for the clarification. This is a nice example.2012-05-12
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    I like this answer.2012-05-12
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    It's interesting...where and how many times have you seen this example used?2012-05-20
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    @rschwieb, I have only ever seen it used as homework for students in the first weeks of a group theory class. I don't know of any non-pedagogic uses.2012-05-20
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    @GerryMyerson I have a feeling it came out of *something* cool and interesting, so I had to ask :)2012-05-20