Proving a set (R, *) is a monoid

Question

For a set to be a monoid, it must be associative and must have an identity element. I've proved that is it associative but don't know how to prove that it has a identity element.

1. Let $\mathbb{R}^3$ be the set of all ordered triples of numbers, and let $\otimes$ be the binary operation on $\mathbb{R}^3$ defined such that $$\left(x_1, y_1, z_1\right) \otimes \left(x_2, y_2, z_2\right) = \left(x_1x_2, x_1y_2+y_1z_2, z_1z_2\right)$$ for all $\left(x_1, y_1, z_1\right), \left(x_2, y_2, z_2\right) \in \mathbb{R}^3$. Prove that $\left(\mathbb{R}^3, \otimes\right)$ is a monoid. What is the identity element of this monoid? Is the monoid $\left(\mathbb{R}^3, \otimes\right)$ a group?

Answer 1

Suppose that there exist a tuple $(x_2,y_2,z_2)$ such that $(x_1,y_1,z_1)\otimes (x_2,y_2,z_2)=(x_1,y_1,z_1)$ for all $(x_1,y_1,z_1)\in \mathbb{R}^3$. Looking at the first coordinate of your formula we get $x_1x_2=x_1$ for all $x_1$ and this implies that $x_2=1$. In the same way, but looking at the third coordinate, we get $z_1z_2=z_1$ for all $z_1$, so that $z_2=1$. Now, it is easy get $y_2=0$. Since the law it is not commutative you need to check the other way around. This is an easy computation and it shows that $(1,0,1)$ is indeed the identity.

This monoid is not a group. For example the element $(0,0,0)$ does not have an inverse because if $(x,y,z)$ were such inverse then $(0,0,0)\otimes (x,y,z)$ would be $(1,0,1)$ but $(0,0,0)\otimes (x,y,z)=(0,0,0)$.

Answer 2

I would try something like $(1,0,1)$ and $(1,1,1)$. In first and third coordinates the operation is just mulitplication, so the unit will be $1$ there.

But the clearest if you write it up:

$(x,y,z)\otimes (1,b,1) = (x,y,z)\ $ and solve it for $b$. You can use specific $(x,y,z)$'s, e.g. plug in $(1,0,0)$, etc.

Answer 3

So for $(e,f,g)$ to be an identity element means in particular that $$(x,y,z)\otimes (e,f,g)=(x,y,z)$$ for all $x,y,z$. Comparing coordinates one gets $xe=x$, $xf+yg=y$ and $zg=z$.

Solving this and checking this and checking which elements satisfy the reverse identity is left as an exercise for you.

Answer 4

Well, you need to find $(x_2,y_2,z_2)$ s.t. $x_1x_2=x_1$, $x_1y_2+y_1z_2=y_1$ and $z_1z_2=z_1$. From this it follows that $x_2=z_2=1$. And $y_2=0$.

Answer 5

The identity element is (1,0,1)

Also $(x,y,z)^{-1}=(x^{-1},\frac{-y}{zx},z^{-1})$, if $x\not=0,z\not=0$

Answer 6

Looks suspiciously like the set of matrices \begin{equation} \begin{bmatrix} x & y\\ 0 & z \end{bmatrix} \end{equation} under matrix multiplication.