Using group presentation for its corresponding semigroup?

Question

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.

Answer 1

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.

Answer 2

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.