Abstract Algebra - Monoids

Question

I'm trying to find the identity of a monoid but all the examples I can find are not as complex as this one.

_{(source: gyazo.com)}

Answer 1

You want a tuple $(a,b,c)$ so that for all tuples $(x,y,z)$, we have $(x,y,z)\otimes (a,b,c) = (x,y,z)$.

By definition, this means that $(xa,xb+yc,zc) = (x,y,z)$, i.e. that $xa = x$, $xb+yc = y$ and $zc = z$ for all $x$, $y$, $z$.

Can you see what this implies about the values of $a$, $b$, and $c$?

Answer 2

Without loss of generality, assume the identity element is $(x_1,y_1,z_1)$. Then, it is clear $x_1=1$ and $z_1=1$. We also know that $x_1y_2+y_1z_2=y_2+y_1z_2=y_2$, which implies $y_1=0$ since $(x_1,y_1,z_1)$ must be the identity for all $x\in\mathbb{R}^3$. Then, we have that the identity is $(1,0,1)$.

Answer 3

It's straightforward to show that $\otimes$ is an associative binary operation, and as others have pointed out, the identity of the monoid is $(1, 0, 1)$. However, $(\mathbb{R}^3, \otimes)$ is not a group, since for example $(0, 0, 0)$ has no inverse element.