# Cancellation in a finite semigroup

1. Jun 13, 2007

So, let G be a nonempty finite set with an associative binary operation such that for all a,b,c in G ab = ac => b = c & ba = ca => b = c, i.e. both left and right cancellation hold. The G is a group.

Ok, I really had no inspiration how to solve this one, so I looked at the solution, which confused me. Namely, it states that, given any element a from G, the set <a> = {a^k : k in N} must be a subset of G, and even better, it is a closed subset of G, and we *consider it in place of G*. This is what confuses me. How can we consider it in place of G? What if there are other elements in G which can not be generated by some a from G? I'm probably missing something simple, but I can't figure it out.

2. Jun 13, 2007

### Dick

{a^k: k in N} is not only closed, it's finite. How can this be? Must be for some k1<k2, a^k1=a^k2. Now think of using your cancellation property. The goal here is to show that 'a' has a inverse.

3. Jun 13, 2007

Yes, it's finite, since it is a subset of a finite set.

Shouldn't I try to show that there exists an identity element in G first?

4. Jun 13, 2007

### Office_Shredder

Staff Emeritus
Yes radou., If ak1=ak2, suppose k2>k1. You can write the right hand side as a multiple of ak1*something if you think about it for a couple seconds.... what is that something?

The trip up is what you just proved exists is dependent on a... so you'll need to prove it's the same one for all elements in G

5. Jun 13, 2007

### Dick

Guess so. I was thinking a semigroup had an identity.

6. Jun 13, 2007

### Office_Shredder

Staff Emeritus
Not knowing the definition of a semigroup off the top of my head, from what you're given you don't know G has an identity (the word semigroup is never used in the problem, so it doesn't even matter that I don't know the definition)

7. Jun 14, 2007

### matt grime

You have a semi group, so you're given things like associativity, so all you need to do is show that you have an identity and an inverse for each element a. To do this it suffices to work in <a> - if we can show it here, it holds for G, since G will be the union of its cyclic subgroups, and all you need to do is show uniqueness of the identity (a priori this construction on cyclic subgroups shows that there are lots of identities).

I don't know why the question bothers to say that, but that is what is going on.