Non-associative, non-commutative binary operation with a identity

Question

Can you give me few examples of binary operation that it is not associative, not commutative but has an identity element?

Answer 1

Here are a couple easy examples, I'll leave it to you to verify their properties.

The natural numbers where $n\ast m=n^m$.

The integers where $n\ast m=n-m$.

The real numbers without zero where $x \ast y = \frac{x}{y}$.

Answer 2

Pick any non-associative, non-commutative operation $\star$ on a set $X$. Now pick a new element $1$ which is not in $X$, let $Y=X\cup\{1\}$, and define a new operation $\bullet$ on $Y$ extending $\star$ and such that $$1\bullet x=x\bullet1=x$$ for all $x\in X$, and $1\bullet1=1$.

Answer 3

The previous examples are all perfectly good, of course, but I thought I would add a rather low-brow way to see that such examples must be easy to generate.

Let $S$ be a finite set, and let's label its elements $x_1,\ldots,x_n$. To define a binary operation $\star: S \times S \rightarrow S$ is equivalent to filling in all the entries of an $n \times n$ matrix $M = \{m_{ij} \}$ with elements of $S$. In particular, since we have $n^2$ entries to fill out and $n$ possible choices for each entry, the total number of such matrices -- i.e., the total number of binary operations on $S$ -- is $n^{n^2}$. This is a pretty big number even when $n$ is rather small.

Suppose we wish to enforce that the first element $x_1$ is a two-sided identity for our operation $\star$. Then what we need is that for all $i,j \in \{1,\ldots,n\}$, $m_{1j} = x_j$ and $m_{i1} = x_i$. In other words, the first row and first column of the matrix are entirely determined, but this leaves us with $n^2 - 2n + 1 = (n-1)^2$ entries left to choose.

It is clear that the operation is commutative iff the matrix is symmetric, i.e., $m_{ij} = m_{ji}$ for all $i,j$. It is easy to give a precise count of the number of symmetric matrices with first row and first column as determined above and even easier to see that, as long as $n \geq 3$, most such matrices are not of this form.

Finally we come to associativity. This is harder to see directly from the matrix, but it is easy to see that we have an awful lot of choices that will result in non-associative operations. For instance, let us say for the sake of argument that $x_2 \star x_3 = x_4$ and $x_3 \star x_4 = x_5$ (all elements distinct): up to relabelling the elements, this is in some sense the generic situation. Then we find

$(x_2 \star x_3) \star x_4) = x_4 \star x_4$

whereas

$x_2 \star (x_3 \star x_4) = x_2 \star x_5$.

Now for a randomly defined binary operation, the chance that these two products are equal is only $\frac{1}{n}$. In any case, we can certainly define them to have different values even among $x_1,\ldots,x_5$ and get a nonassociative binary operation.

If you like counting things, it is a fun exercise to see what you can prove about how many of the $n^{(n-1)^2}$ binary operations on $S$ having $x_1$ as an identity are commutative, or associative, or both, or neither. If $X_n$ is the number of such operations (having $x_1$ as an identity) which are neither commutative nor associative, it should be easy to show that $\lim_{n \rightarrow \infty} \frac{X_n}{n^{(n-1)^2}} = 1$, i.e., that "with probability $1$" a randomly chosen such operation is neither commutative nor associative. I have no doubt that such results appear in the literature, with explicit bounds. If someone knows a nice argument leading to an asymptotic or exact formula for $X_n$, I would be interested to see it.

Answer 4

the octonions! more generally (multiplication in) further algebras coming from the cayley-dickson construction.

Answer 5

Here's an example of an abelian group without associativity, inspired by an answer to this question. Consider the game of rock-paper-scissors: $R$ is rock, $P$ is paper, $S$ is scissors, and $1$ is fold/draw/indeterminate. Let $\ast$ be the binary operation "play".

\begin{array}{r|cccc} \ast & 1 & R & P & S\\ \hline 1& 1 & R & P & S \\ R & R& 1 & P & R\\ P & P& P & 1 & S\\ S & S& R & S & 1 . \end{array}

The mutliplication table above defines a set of elements, with a binary operation, that is commutative and non-associative. Also, each element has an inverse (itself), and the identity exists.

Answer 6

Since the assignment is over, I can send my solution. This is an example: $$ x*y=\begin{cases} x&\text{if }y=0;\\ |x|+y&\text{if }y\not=0. \end{cases} $$

Answer 7

Take a set $S=\{e,\alpha,\beta\}$ with three elements, and define multiplication on $S$ so that $e$ is the identity. To force the multiplication to be non-commutative, ensure that $\alpha\beta\neq\beta\alpha$. To ensure that is non-associative, define $\alpha\alpha=\beta$, so that $a\beta=\alpha(\alpha\alpha)$, and $\beta\alpha=(\alpha\alpha)\alpha$, and hence $(\alpha\alpha)\alpha\neq\alpha(\alpha\alpha)$.

Answer 8

An example of non-associative and non-commutative groupoid with identity is any non-commutative loop and, in particular, this one, finite, of order $12$.