# Proving a set (R, *) is a monoid

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.

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It is not hard to see from the expression that the identity element must be of the form $(1,y,1)$. For what value of $y=y_1$ does one have $1y_2+yz_2=y_2$ for all $(x_2,y_2,z_2)$? Does one also have $(x_1,x_2,x_3)\otimes(1,y,1)=(x_1,x_2,x_3)$ for this value of $y$ and all $(x_1,x_2,x_3)$? –  Marc van Leeuwen Jan 23 '13 at 16:58

## 6 Answers

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 $(0,0,0)\otimes (x,y,z)=(0,0,0)$.

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So does that also mean it's a group? –  Adegoke A Jan 23 '13 at 17:07
To see that it is not a group you can observe that $(0,0,0)\otimes(x,y,z)=(0,0,0)$ for all $(x,y,z)$ so that $(0,0,0)$ does not have inverse. –  Quimey Jan 23 '13 at 17:10
Is there a reason for using (0,0,0)? Why not invert (1,0,1)? –  Adegoke A Jan 23 '13 at 17:13
In a group every element is invertible. The element $(0,0,0)$ is not invertible in your monoid. In the other hand $(1,0,1)$ is his own inverse (and therefore it is invertible$. – Quimey Jan 23 '13 at 17:20 No, It is not a group. I have edited the answer to include these observations. – Quimey Jan 23 '13 at 17:25 Looks suspiciously like the set of matrices $$\begin{bmatrix} x & y\\ 0 & z \end{bmatrix}$$ under matrix multiplication. - 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)$, etc. - Well, you need to find$(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$. - So for$(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. - How did you get$xf + yg = y$? – Adegoke A Apr 10 '13 at 23:43 @AdegokeA This is just comparison of the second coordinate of the result of this "strange" multiplication and$(x,y,z)$. – Julian Kuelshammer Apr 11 '13 at 6:36 But should it not be$xe = x, yf =y$and$zg = z$? – Adegoke A Apr 11 '13 at 10:32 @AdegokeA No, see the definition of$\otimes$in the question. – Julian Kuelshammer Apr 11 '13 at 10:54 It says a binary operation. I thought ⊗ meant multiplication. Those the ⊗ operator have a different function? – Adegoke A Apr 11 '13 at 11:43 The identity element is (1,0,1) Also$(x,y,z)^{-1}=(x^{-1},\frac{-y}{zx},z^{-1})$, if$x\not=0,z\not=0$- This formula makes no sense for$x=0$(or$z\$). –  Quimey Jan 23 '13 at 17:01
Is it really a group? –  Adegoke A Jan 23 '13 at 17:07
If you get rid of the zeros in x and z, then yes. –  Ishan Banerjee Jan 23 '13 at 17:09