# Abstract algebra / binary operation

## Homework Statement

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

## 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.

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member

## The Attempt at a Solution

max(m,0) = always m, so 0 is identity, right??

Yes.

how about unit? 0 is also unit?

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.

in b, if u is a unit, sigma (a) = ua is gonne be identity or a???

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.

Actually, I'm confusing about the concept of unit.

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

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?

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
so..
in b, if u is a unit, sigma(a) = ua can be any number?

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.

or u is inverse of a?

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.