Does commutativity imply associativity? I'm asking this because I was trying to think of structures that are commutative but non-associative but couldn't come up with any. Are there any such examples?
NOTE: I wasn't sure how to tag this so feel free to retag it.
Consider the operation $(x,y) \mapsto xy+1$ on the integers.
A basic example is the "midpoint" binary operation: $a*b = \frac{a+b}{2}$
In general, if $P(u,v)$ is any polynomial in two variables with rational coefficients, then $x*y = P(x+y,xy)$ is rarely associative - I'd be curious under what conditions on $P$ this operation would be associative.
My example is $P(u,v)=\frac{u}{2}$ and Marlu's example is $P(u,v)=1+v$.
Arguably the most important example of a commutative but non-associative structure is that of finite-precision floating point numbers under addition. (a + -a) + b is always equal to b but a + (-a + b) can differ from b since the sum -a + b can involve a loss of precision (this is especially true if a and b are nearly but not quite equal, -a + b could work out to 0 even though the corresponding real sum is nonzero). The lack of associativity of floating point arithmetic is a constant complicating factor in numerical analysis.
(a + -a) + b is also not always equal to b. Consider a = inf, then the result is NaN. Also depending on your definition of 'equality', if a = 0 and b is -0 then the result is 0 - with a different sign than b.
– orlp
Feb 08 '24 at 20:10
Consider the commutative operation $\texttt{vs}$ on the set $\{\textbf{rock}, \textbf{paper}, \textbf{scissors}\}$ abbreviated $\{r,p,s\}$ defined by $$ \begin{array}{c|ccc} \texttt{vs} & r&p&s\\ \hline r & r & p & r \\ p & p & p & s \\ s & r & s & s \end{array}$$ It is not associative since, for example, $$\textbf{paper} \texttt{ vs } (\textbf{scissors} \texttt{ vs } \textbf{rock}) = \textbf{paper}$$ but $$(\textbf{paper} \texttt{ vs } \textbf{scissors}) \texttt{ vs } \textbf{rock} = \textbf{rock}.$$
The easiest Jordan algebra is symmetric square matrices with the operation $$ A \ast B = (AB + BA)/2, $$ similar to a Lie algebra but with a plus sign.
Let $A = \{e,x,y\}$. Define $\cdot$ on $A$ to be $a\cdot e=a$ for all $a$, $e\cdot a= a$ for all a, and $a\cdot b=e$ for all $a$ and $b$ such that $a\neq e$ and $b\neq e$, (i.e. $a,b \in \{x,y\}$).
This operation is commutative, $e$ is the identity, (everything even has an inverse), but is not associative since $(x \cdot y) \cdot y = e \cdot y = y$ and $x \cdot (y \cdot y) = x \cdot e = x$.
The simplest examples of commutative but nonassociative operations are the NOR and NAND operations (joint denial and alternative denial) in propositional logic. Quoting from my answer to the question A conjecture in equational logic:
Namely, the $2$-element structure $\{a,b\}$, where $aa=b$ and $ab=ba=bb=a$, is commutative but not associative; in fact, for any $x\in\{a,b\}$, we have $$(ax)b=b(xa)=a,$$ $$a(xb)=(bx)a=b.$$ This is the unique (up to isomorphism) binary operation on a $2$-element set which is commutative but not associative; it can be interpreted as either of the truth-functions NOR or NAND.
The NAND is commutative but not associative.
\begin{eqnarray*} \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline A & B & & A \text{ nand } A & (A \text{ nand } A) \text{ nand } B & & A \text{ nand } B & A \text{ nand } (A \text{ nand } B) \\ \hline 0&0& &1&1& &1&1 \\ \hline 0&1& &1&0& &1&1 \\ \hline 1&0& &0&1& &1&0 \\ \hline 1&1& &0&1& &0&1 \\ \hline \end{array} \end{eqnarray*}
The arithmetical, geometrical and harmonic mean operations on $\mathbb{R}$, $\mathbb{R^+}$ and $\mathbb{R_*^+}$ resp.:
$$a*b=\dfrac{a+b}{2}$$
$$a*b=\sqrt{ab}$$
$$a*b=\dfrac{ab}{a+b}$$
Remark: the non-associativity of arithmetical mean has a kind of physical interpretation by placing more weight either on the last or on the first term:
$$(a*b)*c=\dfrac{\tfrac{a+b}{2}+c}{2}=\dfrac{a+b+2c}{4}$$
whereas
$$a*(b*c)=\dfrac{2a+b+c}{4}$$
(none of them being equal to $\dfrac{a+b+c}{3}$ !...)
For $x,y \in \mathbb{Z}$, define $x*y = xy + 1$.
Then * is clearly commutative.
As for associativity, $$a*(b*c) = a*(bc + 1) = a(bc + 1) + 1 = abc + a + 1$$ $$(a*b)*c = (ab + 1)*c = (ab + 1)c + 1 = abc + c + 1$$ so associativity fails for any triple $(a,b,c)$ with $a \ne c$.
Here's another example . . .
For $x,y \in \mathbb{Z}$, define $x y = x^2y^2$.
Once again, commutativity is obvious.
For associativity, $$a*(b*c) = a*(b^2c^2) = a^2(b^2c^2)^2 = a^2b^4c^4$$ $$(a*b)*c = (a^2b^2)*c = (a^2b^2)^2c^2 = a^4b^4c^2$$ so associativity fails if $a,b,c \ne 0$, and $|a| \ne |c|$.
One last example . . .
For $x,y \in \mathbb{Z}$, define $x*y = -x-y$.
Commutativity is clear.
For associativity, $$a*(b*c) = a*(-b-c) = -a-(-b-c) = -a + b + c$$ $$(a*b)*c = (-a-b)*c = -(-a-b)-c = a + b -c $$ so associativity fails if $a \ne c$.
Using mace4 on the assumption x*y=y*x. and goal (x*y)*z=x*(y*z). gives:
% Interpretation of size 2
*(0,0) = 1.
*(0,1) = 0.
*(1,0) = 0.
*(1,1) = 0.
c1 = 0.
c2 = 0.
c3 = 1.
i.e. 0*0=0, 0*1=0, 1*0=0, 1*1=0, with (x,y,z)=(0,0,1) being the counter-example.
(0*0)*1 = 1*1 = 0
0*(0*1) = 0*0 = 1
P.S. Mace4 is a software used for constructing finite models and finding counter-examples.
Given ideals $I$ and $J$ in a Lie algebra $\mathfrak{g}$ over a commutative ring $K$, the ideal product $[I,J]$ defined as the image of the composite $$I \otimes_K J \hookrightarrow \mathfrak{g} \otimes_K \mathfrak{g} \xrightarrow{[-, -]} \mathfrak{g}$$ is commutative, but not associative in general. It is easy to see why, because the ideal product is given by $$[I,J] \equiv \{ \Sigma_{i=0}^n [x_i, y_i] \mid \text{$x_i \in I$, $y_i \in J$, $n \in \mathbb{N}$} \},$$ and $[x,y]=-[y,x]$, but the bracket is not associative in general.
For our example the carrier set will be the set ${\Bbb Z}^{\Bbb Z}$ of all functions from $\Bbb Z$ to $\Bbb Z$.
For any two function $f, g \in {\Bbb Z}^{\Bbb Z}$ we can form the pointwise sum, $f + g: n \mapsto f(n) + g(n)$.
We have the (left) bilateral shift operator $T$ on ${\Bbb Z}^{\Bbb Z}$,
$\tag 1 T: (a_k)_{k=-\infty}^\infty \mapsto (a_{k+1})_{k=-\infty}^\infty$
We define a binary operation $\oplus$ on the set ${\Bbb Z}^{\Bbb Z}$ as follows,
$\tag 2 (f,g) \mapsto f \oplus g = T \circ (f+g)$
This binary operation is commutative but not associative; it satisfies
$\tag 3 f \oplus g = h \oplus k \; \text{ iff } \; f + g = h + k$
