# Abstract algebra / binary operation

1. Feb 16, 2009

### hsong9

1. The problem statement, all variables and given/known data
a. In each case a binary operation * is given on a set M. Decide whether it is commutative or associative, whether an identity exists, and find the units.
M=N(natrual); m*n = max(m,n)

b. If M is a moniod and u in M, let sigma: M -> M be defined by sigma(a) = ua for all a in M.
(a) show that sigma is a bijection if and only if u is a unit.
(b) If u is a unit, describe the inverse mapping sigma^-1: M -> M

2. Relevant equations
3. The attempt at a solution
In a, I know it is commutative and associative. I'm not sure identity and unit.
max(m,0) = always m, so 0 is identity, right?? how about unit? 0 is also unit?

in b, if u is a unit, sigma (a) = ua is gonne be identity or a???
Actually, I'm confusing about the concept of unit.

Thanks.

2. Feb 16, 2009

### Tom Mattson

Staff Emeritus
Yes.

Ask yourself, "Does 0 have an inverse in this set?" That is, "Does there exist an element x in this set such that x*0=0?" I think the answer is pretty clear.

Why does it have to be either? A unit is just an invertible element in a set. For instance consider the real numbers under multiplication. All nonzero elements are units. So let a=5 and u=2. Then $\sigma (5)=(2)(5)=10$ which is neither the identity nor a.

That's what definitions are for. Have you read the definition of "unit"?

3. Feb 16, 2009

### hsong9

so..
in b, if u is a unit, sigma(a) = ua can be any number?
or u is inverse of a?
if u is inverse of a, sigma(a) = ua = 1..
this is not bijection.. right?

4. Feb 16, 2009

### Tom Mattson

Staff Emeritus
Well, $a$ can be any element of the monoid. The proposition that $ua$ can be any element is one of the things you're supposed to prove.

Not necessarily. In fact, since $a$ need not be a unit there's no reason to think that $a$ even has an inverse.

You seem to be assuming several things that are not implied by the problem statement. You really have to focus only on what is written there, and don't add anything to it.