Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Homework

share|improve this question

3 Answers 3

up vote 7 down vote accepted

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 $c$?

share|improve this answer
    
No, sorry. I've only started abstract algebra so I don't know what it implies. –  Adegoke A Jan 4 '13 at 0:29
2  
Abstract algebra is not needed to complete Alex Kruckman's argument. For example, since he showed that $xa=x$ for all $x$, you can determine $a$ by elementary algebra. Then you can similarly determine $c$ and finally $b$. –  Andreas Blass Jan 4 '13 at 0:55
    
OK. Now I understand. Thanks a lot! –  Adegoke A Jan 4 '13 at 1:13

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 $(1,0,1)$.

share|improve this answer
    
Great answer but what allows us to "assume the identity element is (x1,y1,z1)" ? Also you said " it is clear x1=1 and z1=1" how did you come to this conclusion? –  Adegoke A Jan 4 '13 at 0:34
2  
@AdegokeA The first part is just a heuristic: to determine what the element would be if it existed; once you have determined sufficient information about it to construct the element you go back and check that it works. The second part is by examination of the first and third components, here a and b, $(a,.,b)$. –  peoplepower Jan 4 '13 at 0:37

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.

share|improve this answer
    
Is there any reason for using (0,0,0) as an example? –  Adegoke A Jan 4 '13 at 1:18
    
@Adegoke A Any element of the form $(0, a, b)$ has no inverse element. –  Vectk Jan 4 '13 at 1:28
    
OK. But what is being inverted that leads to the conclusion that it's not a group? –  Adegoke A Jan 4 '13 at 1:33
    
@AdegokeA A group is a monoid with the additional requirement that every element has an inverse. Since $(0,a,b)$ has no inverse, $(\mathbb{R}^3, \otimes)$ is not a group. –  Vectk Jan 4 '13 at 1:38
    
Yes, I understand that. I just don't know where (0,a,b) came from. –  Adegoke A Jan 4 '13 at 1:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.