0
$\begingroup$

Somewhere Colin M. Campbell noted:

If $A$ is a semigroup defined as $$A=Sg(\pi)=\langle a_1,\cdots, a_d\mid u_1=v_1,\cdots,u_e=v_e\rangle $$ then the same generators with the same relations can also be interpreted as the presentation of the following group: $$A^*=Gp(\pi)=\langle a_1,\cdots, a_d\mid u_1=v_1,\cdots,u_e=v_e\rangle $$

So he consider a semigroup first and then its possible analogous group form. I am working on this ideas and asking if we can do converse direction? I mean can we start with a (for example) finite group such as $$D_8=\langle x,y\mid x^2=y^4=(xy)^2=1\rangle$$ and then construct a finite semigroup accordingly? I see, I can write $x^2=1$ as $x^3=x$ without any problems but I want to be sure about doing this job. Sorry if this question seems ridiculous for you. Thanks for your comments and your time.

Edit: By saying corresponding semigroup, I mean one possible related semigroup.

$\endgroup$

2 Answers 2

4
$\begingroup$

You can do this in some examples. For example in braid groups or, more generally, Artin groups, you can just interpret the group presentation immediately as a semigroup (or monoid) presentation. In general, in that situation, the semigroup defined will not embed into the group. An obvious example is $\langle x \mid x^4=x^2 \rangle$, which defines a semigroup of order $3$ and a group of order $2$. It turns out that, for Artin groups, the semigroup (or monoid) does embed into the group, but that is a nontrivial result.

If $1$ occurs in the presentation, as in your example, then you can still interpret it as a monoid presentation, and it is easy to see that the monoid defined by $\langle x,y \mid x^2 = y^4 = (xy)^2 = 1 \rangle$ is the same as the group. That is essentially because the first two relations tell you that $x$ and $y$ have inverses.

But, in general, the relations of a group presentation, such as the surface group $\langle w,x,y,z \mid wxw^{-1}x^{-1}yzy^{-1}z^{-1} \rangle$ will contain inverses of generators, and I don't see any way that you can make a semigroup or monoid out of that, unless you want to adjoin the inverses as extra generators.

$\endgroup$
0
3
$\begingroup$

One problem with your proposal is that, although every semigroup uniquely determines a group (by formally adjoining an identity element and inverses), it is not true that every group uniquely determines a semigroup in the way that you suggest. Even in your example, where you suggested replacing $x^2=1$ with $x^3=x$ (to avoid mentioning $1$), you could also have used $x^2y=y$ or $x^{10}=x^8$ or any combination of such things, since they all become equivalent to $x^2=1$ in groups. So it's not clear what you mean by "its corresponding semigroup" in the title of your question; many semigroups can lead to the same group.

$\endgroup$
1
  • $\begingroup$ It is definitely true that a group could make many semigroup as you noted. Thanks for making clear. Indeed, I meant one possible semigroup. So I can do what I was looking for. $\endgroup$
    – Mikasa
    Aug 5, 2014 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.