Can a Semigroup with Finitely Many Elements and the Cancellation Law be a Group?

[T-O7]

Hey,
I'm having trouble on a question about semigroups which looks like it should be fairly straightforward:
If S is a semigroup with finitely many elements satisfying the Cancellation Law (i.e. ab=ac implies b=c and ba=ca implies b=c), then S is a group.

Anyone know how to go about this?

 

[dogma]

don't know if this will help, but...

if $$a b = a c$$ where $$a \neq 0$$

$$a b - a c = 0$$

$$a (b - c) = 0$$

since $$a \neq 0$$, $$b - c = 0$$

and then $$b = c$$


I hope this is somewhat useful (and correct )

Best!

 

[matt grime]

sadly, dogma, this isn't a question about rings, and so one may not add elements like that, so strictly speaking it is not useful. nor is it correct since even if this were a question about rings it isn't specified that there are no zero divisors.

since the semi group is finite, you can show that the map $$f_x: a \to ax$$
induces a map to some general linear group, that shows the semigroup is also a group.

MOre directly, by repeated use of the pigeon hole principle using such maps, you may demonstrate that some f_x is the identity mapping, that inverses exist and you are done.

 

[NateTG]

sadly, dogma, this isn't a question about rings, and so one may not add elements like that, so strictly speaking it is not useful. nor is it correct since even if this were a question about rings it isn't specified that there are no zero divisors.

since the semi group is finite, you can show that the map $$f_x: a \to ax$$
induces a map to some general linear group, that shows the semigroup is also a group.

MOre directly, by repeated use of the pigeon hole principle using such maps, you may demonstrate that some f_x is the identity mapping, that inverses exist and you are done.

The original question was ambiguous about whether the semigroup contains finitely many elements (a finite semigroup) or the semigroup has a finite subset that satisfys the cancellation law.

An alternative approach would be to use a 'semi-group action':
Let's say the semi-group is $$H$$ and $$|H|=n$$ since $$H$$ is finite. Then left multiplication by elements in $$H$$ can be considered to be a permutation of the elements of $$H$$ since semi-groups are closed over multiplication, and the cancellation rule forces the operation to be bijective. This means that $$H$$ is isomorphic (I think that's the right word) to a subset of the elements of the permutation group of $$n$$ elements - $$S_n$$, call it $$S$$. Then $$H$$ must also be isomorphic to the subgroup generated by $$S$$ since both are generated by multiplication.

 

[T-O7]

Thanks NateTG, I realized that I had interpreted the ambiguous question incorrectly, and thought it meant "the semigroup has a finite subset that satisfies the cancellation law", when now I'm pretty the question is actually stating "the semigroup contains finitely many elements, ALL of which satisfy the cancellation law".

With the latter interpretation, I used your train of thought and showed that the "cosets" $$aS$$ of the semigroup $$S$$ was the semigroup itself ($$a\in S$$), and that consequently each element (using the cancellation law too) had an inverse in $$S$$. I assumed the semigroup $$S$$ contains the identity, something that doesn't seem to occur in the usual definition of a semigroup, but is somehow given with it in my textbook.

Thanks everyone.

 

[matt grime]

I read it is a semigroup with finitely many elements, and the semigroup has the cancellation property... since the most of the other interpretations are trivially false.