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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. –  Xabier Domínguez May 11 '12 at 15:41
    
Oh man, I hadn't even thought about associativity! –  MathematicalOrchid May 11 '12 at 15:45
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;) You cannot go very far without it. –  Xabier Domínguez May 11 '12 at 15:45
    
what happens if you consider ${0,1}$ as the candidates . –  Theorem May 11 '12 at 15:46
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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. –  anon May 11 '12 at 15:49

2 Answers 2

up vote 3 down vote accepted

$x^y$ 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? –  MathematicalOrchid May 11 '12 at 15:47
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Well, if you restrict it to the set $\{1\}$… –  MJD May 11 '12 at 16:22

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. –  chris May 12 '12 at 8:30
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@chris the base of the logarithm. If you chose $\ln$ it will be $e$. –  Asaf Karagila May 12 '12 at 8:33
    
I realized after a bit more thought my earlier comment was incorrect. Thanks for the clarification. This is a nice example. –  chris May 12 '12 at 8:38
    
I like this answer. –  MathematicalOrchid May 12 '12 at 12:46
    
It's interesting...where and how many times have you seen this example used? –  rschwieb May 20 '12 at 16:52

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