semi group with property is a group question

k. can't really get anywhere with this.

given a semi group (associativity and closure on an operation defined on a set) S

Show that if S has the property that for all

\displaystyle x \in S \exists ! x'\in S

s.t.

xx'x=x

show that S is a group.

Now x'=x^-1 and xx' is the identity but I can't seem to get this.

I was trying to show that (ab)'=b'a' and (a')'=a but can't seem to get them with pencil pushing. Any hints?

(edited 12 years ago)

Reply 1

nuodai

You're right that you need to show that

xx'

is an identity for any

x

and that

x'=x^{-1}

. But you can't use results like cancellation or inversion to show this since they rely on the assumption that an inverse for each element exists, which is what you're trying to prove.

For simplicity's sake, let

e_x = xx'

. We know that

e_xx=x

, so you need to show that

xe_x=x

, and perhaps that

e_x=e_y\, (=e)

for all

x,y

. I imagine uniqueness will be quite important here. You can probably use some sort of symmetry argument to show that

x'x=e

, in which case the fact that

x'=x^{-1}

is immediate.

I haven't actually tried this myself; these are just my thoughts.

EDIT: In fact, showing that

xx'=x'x

and

xxx'=x

is sufficient, since an identity (if it exists) must be unique. [If

e_1, e_2

are two identities, then

e_1=e_1e_2=e_2

(edited 12 years ago)

Reply 2

SimonM

Not really. We also need yxx' = y.

Reply 3

DFranklin

nuodai

I agree using uniqueness is going to be important, but I'm not getting anywhere in terms of showing x'x = xx' or similar. (The most I've managed is showing x'xx' = x', so x'' = x).

My minds definitely not "thinking the right way" for this one.

Reply 4

Farhan.Hanif93

It's quite likely I'm out of my depth on this one but is it not sufficient to show that the idempotents of S commute as this, coupled with the fact that S is a regular semigroup (by definition), would mean that each element has a unique inverse?

As the others have pointed out, first show that xx'x=x yield x'xx'=x' and let u,v in S be two idempotents with x=(uv)^-1. Use this to show that they commute.

Then assume x has inverses x' and x'' in S so:

x=xx'x
x'=x'xx'
x=xx''x
x''=x''xx''

xx', x'x, xx'', x''x are all idempotents (that commute if you show the first thing I mentioned) so you can show that xx'=xx'' and analogously x'x=x''x.

Using all these results, you can show that x'=x'' and hence each element has a unique inverse.

Not certain on how I'd show that there's a unique identity but this should be a decent start.
EDIT: Any confirmation that I'm not speaking rubbish here, would be good.

(edited 12 years ago)

Reply 5

DFranklin

I don't see how you can say x = (uv)^{-1} when we don't know that inverses exist.

Reply 6

Mark13

Original post by Farhan.Hanif93

Using all these results, you can show that x'=x'' and hence each element has a unique inverse.

x' = x'' only if x has order 1 or 2.

EDIT: Just realised that you're not using the notation x' in the same way as the OP, my bad.

(edited 12 years ago)

Reply 7

Farhan.Hanif93

Original post by DFranklin

I don't see how you can say x = (uv)^{-1} when we don't know that inverses exist.

Doesn't the following show that inverses exist for each element:

x(x'xx')x=xx'(xx'x) = xx'x = x

shows that x has inverse x'xx' in S?

(edited 12 years ago)

Reply 8

Farhan.Hanif93

Original post by Mark13

x' = x'' only if x has order 1 or 2.

EDIT: Just realised that you're not using the notation x' in the same way as the OP, my bad.

I should have been a bit more clear with that. My apologies. :p:

Reply 9

Zhen Lin

Original post by Totally Tom

k. can't really get anywhere with this.

given a semi group (associativity and closure on an operation defined on a set) S

Show that if S has the property that for all

\displaystyle x \in S \exists ! x'\in S

s.t.

xx'x=x

show that S is a group.

Observe that x(x'xx')x = xx'x by associativity, so by uniqueness, x' = x'xx'. Therefore we have an inverse semigroup. But there are inverse semigroups which aren't groups, such as the bicyclic semigroup. Are you sure the question is correct?

Reply 10

Mark13

Original post by Farhan.Hanif93

Doesn't the following show that inverses exist for each element:

x(x'xx')x=xx'(xx'x) = xx'x = x

shows that x has inverse xx'x in S?

You can't show that any element has an inverse until you know there is an identity element. Assuming that you're using x' in the sense of the OP, remember that xx'x = x, and so in general will not be the inverse of x.

Reply 11

DFranklin

Original post by Farhan.Hanif93

Doesn't the following show that inverses exist for each element:

x(x'xx')x=xx'(xx'x) = xx'x = x

shows that x has inverse x'xx' in S?

What's your definition of inverse?

Reply 12

Farhan.Hanif93

Original post by Mark13

Original post by DFranklin

What's your definition of inverse?

Fair point. I think I'll leave this to you guys with more experience. :o:

Reply 13

Farhan.Hanif93

Original post by Mark13

I know I said that I would leave this to you guys but I've just done a little reading up which suggests that, in the context of a semigroup, it is possible to define an element x' as the inverse of x if xx'x=x and x'xx'=x'.

I think Zhen has a point, though. The question appears to be incorrect unless we're missing some information like x' is unique for each x (i.e. the existence of a unique pseudoinverse).

EDIT: just realised I misunderstood the notation '!'. The question seems right in this case since, if we can show that the semigroup is inverse, then the existence of a unique pseudoinverse implies the existence of a unique inverse for each element in S.

(edited 12 years ago)

Reply 14

DFranklin

A little help from Farhan led me to an actual posted proof: http://planetmath.org/?op=getobj&from=objects&id=6391

I'm not sure at what point you should 'give up and refer to it'. I knew nothing about semigroups except what was in the original post, and I think it's a pretty tough ask on that basis (e.g. if you were doing a groups example sheet and a question says "A semigroup is a set and operation ... Suppose a semigroup S has the property ...; show that S is in fact a group", then that is a pretty mean question). If you'd done some other stuff on semigroups, maybe you should think about it a bit more before referring to the proof.

Reply 15

Totally Tom

Thanks for that link. I've got no experience with semigroups (other than the definition). The question comes from "A course on Finite Groups" by Harvey E. Rose. Good book aside from awful notation.