# Alternative group laws over the reals

I am tutoring a younger student in Algebra (primary school) and during the discussion about laws of addition and multiplication (commutativity, associativity, distributivity) I mentioned to her that a lot of the things we are learning about (polynomials, solving equations,etc.) apply equally if we use completely different operations to combine numbers as long as they obey the same set of rules. She seemed to buy this, but did not look satisfied without an example - and I wasn't about give a 13 year old a crash course in abstract algebra to use a more complicated example.

Later when I got home I realized that I didn't know of the existence of any group laws over the set of real numbers other than the canonical addition and multiplication laws. Do any exist? I would love to present something like this as a problem at our next lesson.

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Over the reals may be difficult for a natural not too technically complicated example. What about the good old clock arithmetic, integers $0$ to $11$ under addition modulo $12$? –  André Nicolas Jun 22 '11 at 6:48
I agree with user6312, modular arithmetic will probably give easier examples of different structures on the same set. You could do $\mathbb{Z}/4\mathbb{Z}$ vs. $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ maybe. Another possibly easier example for your student, if you have shown $\mathbb{Q}$ is countable: $\mathbb{Z}$ vs. $\mathbb{Q}$. –  Zev Chonoles Jun 22 '11 at 6:59

A standard exercise is to show that $a!b=ab+a+b$ makes a group out of the reals, omitting $-1$.

EDIT: Remembered another example: $a\star b=\root3\of{a^3+b^3}$ makes a group out of the reals.

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And is it also standard to include the "debriefing" as to where this crazy operation comes from? (Hint for those who don't know: compare this to $x \cdot y = xy$: the latter has an identity element of $1$ and behaves badly at zero, whereas Gerry's group law...) –  Pete L. Clark Jun 22 '11 at 7:23

If we only need to use the real numbers as a set, it is relatively easy to construct such examples (although I don't think they'll be particularly illuminating); one example would be, the group structure of a free abelian group with an uncountable basis, e.g. $\bigoplus_{\mathbb{R}}\mathbb{Z}$. This set has the same cardinality as $\mathbb{R}$, and choosing some bijection, we can give $\mathbb{R}$ that group structure; but it is a different group structure than that of $\mathbb{R}$ (see here for an explanation of why $\mathbb{R}$ can't be a free abelian group). Note that the free group on $\mathbb{R}$ also has the same cardinality as $\mathbb{R}$, but as it's nonabelian I felt that that might be harder to grasp.

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Why the downvote? –  Zev Chonoles Jun 22 '11 at 6:56
@Zev: I have no idea, but +1 from me. Note that another way to phrase the answer is via a few magic words which came up recently on this site: it is enough to use transport of structure from any group $G$ of continuum cardinality which is not isomorphic to $(\mathbb{R},+)$. –  Pete L. Clark Jun 22 '11 at 7:17
I didn't downvote, but I suspect if you tried to tell a primary school student about a "free abelian group with an uncountable basis," and that student isn't Terry Tao, you won't accomplish very much. –  Gerry Myerson Jun 22 '11 at 7:31
@Gerry: I wouldn't myself call a 13-year old in "primary school", but your point is well-taken: I certainly wouldn't discuss any of these things with a junior high school student and probably not even a high school student, however bright. (For this much abstraction there are many things that I find more contentful...) But I think the pedagogy is clearly part of the setup/background of the question and not the question itself, so there's probably no need to criticize or defend it. –  Pete L. Clark Jun 22 '11 at 19:28
Whoever keeps flagging Gerry's post needs to cut it out. I am not deleting it. It was clearly not meant to be taken either literally or at face value. –  Qiaochu Yuan Jun 25 '11 at 4:40

One can present the usual laws but in disguise. For example, $a\ast b=a\sqrt{1+b^2}+b\sqrt{1+a^2}$ is based on a one-to-one function $u$ on the real line in the sense that $u(a\ast b)=u(a)+u(b)$. Any such function $u$ gives rise to a group structure.

Debriefing (in the sense of Pete L. Clark): $u=\sinh^{-1}$.

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right. As you figured out, I was thinking along the same lines. I almost answered the question by saying "Choose a bijection $\sigma: \mathbb{R} \rightarrow \mathbb{R}$ and conjugate the usual group law by $\sigma$ to get a new one." The only reason I didn't was that it is possible that this gives the original group law, albeit "very unlikely" in the sense that the set of all such $\sigma$ must be very small in the set of all bijections of $\mathbb{R}$. And then I stopped thinking about it. I wonder if anyone knows of a way to formalize this? –  Pete L. Clark Jun 22 '11 at 14:31
Unless I am mistaken, you ask the functions $u$ for which $a\ast b=a+b$, that is, $u(a+b)=u(a)+u(b)$ for every $a$ and $b$. Well, we know the answer, don't we? –  Did Jun 22 '11 at 14:45
thanks, that's a good observation. So in particular we may conjugate by any measurable but nonlinear bijection to get an isomorphic, but different, group law on $\mathbb{R}$. –  Pete L. Clark Jun 22 '11 at 19:25