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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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    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. $\mathbb{Q}$.2011-06-22

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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...)2011-06-22
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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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    Whoever keeps flaggi$n$g 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.2011-06-25
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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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    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 $\mathbb{R}$.2011-06-22