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When I read Pinter's A Book of Abstract Algebra, Exercise 7 on page 25 asks whether the operation $$x*y=\frac{xy}{x+y+1}$$ (defined on the positive real numbers) is associative. At first I considered this to be false, because the expression is so complicated. But when I worked out $(x*y)*z$ and $x*(y*z)$, I found both to be $$\frac{xyz}{xy+yz+zx+x+y+z+1}!$$ Commutativity is easy to see. But associativity can be so counter-intuitive! Can you see this operation is associative without working it out? Are there tricks to do this?

Arnaud D.
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Zirui Wang
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2 Answers2

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A common way to build weird-looking associative operations is to start from a known one, such as multiplication, say on the real numbers or some subset of them, and then to transform it through some bijection $\alpha$, by defining $$x\ast y=\alpha^{-1}(\alpha(x)\cdot\alpha(y)).$$ Indeed this is equivalent to $\alpha(x\ast y)=\alpha(x)\cdot \alpha(y)$ (so that $\alpha$ is actually an isomorphism), and it is then easy to check associativity by noticing that \begin{align*}\alpha(x\ast (y\ast z)) & =\alpha(x)\cdot \alpha(y\ast z) = \alpha(x)\cdot(\alpha(y)\cdot \alpha(z))\\ & =(\alpha(x)\cdot \alpha(y))\cdot \alpha(z) = \alpha(x\ast y)\cdot \alpha(z)\\ & =\alpha((x\ast y)\ast z),\end{align*} which implies that $x\ast (y\ast z)=(x\ast y)\ast z$ since $\alpha$ is bijective. Other properties, such as commutativity or existence of neutral or inverses, can be done in the same way, depending on the cases.

In this case, we can see that $$\frac{1}{x\ast y}=\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$$so that $$1+\frac{1}{x\ast y}=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\left(1+\frac{1}{x}\right)\cdot\left(1+\frac{1}{y}\right),$$ so if you define $\alpha(x)=1+\frac{1}{x}$, you can check that it defines a bijection $(0,+\infty)\to (1,+\infty)$, and $\ast$ is just a transformation of the multiplication on $(1,+\infty)$, which explain why it is associative. In fact you can also see right away that it must also be commutative, but that it can't have a neutral element (otherwise $(1,+\infty)$ would have one).

Arnaud D.
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    A natural follow-up question: Is there an associative operation $$ on an interval $I$ which is not* of this form, i.e. for which there does not exist a function $\alpha:I\to\mathbb R$ such that $\alpha(x*y)=\alpha(x)\alpha(y)$? –  Feb 19 '18 at 10:33
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    @Rahul As I said, any operation constructed this way must be commutative as well, so any non-commutative operation would satisfy your requirement. A nice example of such an operation is to define $x\ast y=y$. This is defined on any interval, associative, but not commutative, and thus it does not come from a "transformed" multiplication. – Arnaud D. Feb 19 '18 at 10:43
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    @Rahul But before you ask, commutativity is not sufficient either : on any subset of $\mathbb{R}$ you can also define $x\ast y = \max{x,y}$. This is commutative and associative, but not a transformed multiplication either because $x\ast y$ must always be either $x$ or $y$. – Arnaud D. Feb 19 '18 at 10:47
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    Thanks for the nice couple of counterexamples. Interestingly, $\max{x,y}$ can be approximated arbitrarily well using $\alpha(x)=\exp(\exp(kx))$ as $k\to\infty$. –  Feb 19 '18 at 11:03
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    @Rahul: your question is basically asking whether or not there is exactly one group of cardinality continuum (up to a unique isomorphism). The answer is a very definite no. There are at least $\mathfrak c$ many non-isomorphic abelian groups of that cardinality (just consider all possible infinite products of groups of the form ${\bf Z}/p{\bf Z}$). There are actually $2^{\mathfrak c}$ many (by instability), though I don't see that many concrete examples at a glance. $SL_3(K)$ with $K$ ranging over all fields of cardinality $\mathfrak c$ should suffice. – tomasz Feb 19 '18 at 15:24
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    Can they all be achieved as limits of that form? I.e. in some sense are they dense? – R.. GitHub STOP HELPING ICE Feb 19 '18 at 21:57
  • @tomasz what do you mean by "by instability?" I've looked it up but I can't find references to it outside of this thread in the context of abstract algebra. – exists-forall Feb 20 '18 at 09:44
  • @tomasz you don't even need to go to groups, this question is only about monoids, which can be more pathological. But, interesting that there are already $2^\mathfrak{c}$ counterexample groups, I wouldn't have expected that. – leftaroundabout Feb 20 '18 at 09:55
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    @SelectricSimian: I mean that there are unstable (in the model-theoretic sense) groups. An unstable countable theory automatically has a maximal number of non-isomorphic models in every uncountable cardinality. I would not be surprised if there was a maximal numbers of abelian groups, but I don't see any argument for that. – tomasz Feb 20 '18 at 21:07
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Another way to prove associativity quite easily is when your operation is already known to be commutative. Then it is enough (and in fact necessary) to find an expression $$ (x\ast y) \ast z = \varphi(x,y,z) $$ where $\varphi$ is invariant under (circular) permutation, meaning that $\varphi(t_1,t_2,t_3) = \varphi(t_{\sigma(1)},t_{\sigma(2)},t_{\sigma(3)})$ for all (circular) permutations $\sigma \in \mathfrak S_3$.

Indeed, you then get $$ (x\ast y) \ast z = \varphi(x,y,z) = \varphi(y,z,x) = (y\ast z)\ast x = x \ast (y \ast z) $$ where the last equality holds by commutativity of $\ast$. (Taking a good look at what happen above, showing that $\varphi$ is invariant under the transposition $(1\,3)$ is actually enough.)

In you case, commutativity of $\ast$ is directly given by commutativity of product and sum of reals. And the expression you found for $(x \ast y)\ast z$ is clearly invariant by permutation of $x,y,z$ (again by commutativity of products and sums of reals). So you can conclude that $\ast$ is associative.

Pece
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