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I am sorry if my question is too simple.

Is every semigroup associated to a group? If no, what conditions should be satisfied for a semigroup to have an associated group? If yes, how can I find the group?

I thought of the universal property. Let $X$ be a semigroup, define $G$ to be the group with a morphism of semigroups $\tau: X \rightarrow G$, such that if $H$ is any group having similar morphism $\phi: X \rightarrow H$, there is a group homomorphism $\psi: G \rightarrow H$, such that $\phi = \psi \circ \tau$.

Uniqueness is obtained from the universal property. But existence needs construction. When there is no element $x$ in $X$ such that for any $a \in X$, $xa = ax = a$, i.e., $X$ doesn't contain an identity element, I define $G = \{ 1 \} \cup \{ x_1^{k_1} \cdots x_n^{k_n} | x_i \in X, k_i \in \{ \pm 1 \}, i = 1, \cdots, n; k_jk_{j+1} = -1, 1 \leq j <n \}$ with equivalence relations: $x^{k}x^{-k} \equiv 1$, $x_1^{k_1} \cdots x_{i-1}^{k_{i-1}}x_i^kx_i^{-k}x_{i+2}^{k_{i+2}} \cdots x_n^{k_n} \equiv x_1^{k_1} \cdots x_{i-1}^{k_{i-1}}x_{i+2}^{k_{i+2}} \cdots x_n^{k_n}$ and group operations

Multiplication: $\begin{cases} 1 \cdot x_1^{k_1} \cdots x_n^{k_n} = x_1^{k_1} \cdots x_n^{k_n} \cdot 1 =x_1^{k_1} \cdots x_n^{k_n}, \\ x_1^{k_1} \cdots x_n^{k_n} \cdot y_1^{l_1} \cdots y_m^{l_m}= \begin{cases} x_1^{k_1} \cdots x_n^{k_n}y_1^{l_1} \cdots y_m^{l_m} & \text{ if } k_nl_1=-1, \\ x_1^{k_1} \cdots (x_ny_1)^{l_1} \cdots y_m^{l_m}& \text{ if } k_nl_1=1; \end{cases} \end{cases}$

Inverse: $1^{-1} =1$, $(x_1^{k_1} \cdots x_n^{k_n})^{-1} = x_n^{-k_n} \cdots x_1^{-k_1}$.

Then clearly, the morphism $\tau: X \rightarrow G$ can be defined to send every $x \in X$ to $x \in G$.

Beside possible leaks, this construction is not precise, and doesn't work in case when $G$ already has an identity, or even is a group.

So, when are semigroups associated with a group, and how can I construct a group from a semigroup?

Thanks in advance.

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    $\begingroup$ Your universal property tells you how to define $G$. Let $G$ be the free group generated by the elements of $X$ modulo the normal subgroup generated by the relations given by the multiplication table in $X$. By design this has the universal property you want. This is called the group completion. $\endgroup$ Oct 30, 2011 at 19:12
  • $\begingroup$ @Ryan Budney: Thank you very much for the comment. So $G$ has the presentation $\langle X | R \rangle$, with $X$ the underlying set of the semigroup and $R$ the multiplication table. $\endgroup$ Nov 3, 2011 at 15:41
  • $\begingroup$ Your question is correct. I refer you to the paper Special Algebraic Structures by Florentin Smarandache, University of New Mexico. This paper contains an example for this semigroup contains a group. $\endgroup$ Aug 2, 2013 at 7:14
  • $\begingroup$ Please don't use capitalisation throughout. Also, could you say a couple of words on what the reference actually does? It isn't immediately clear from your answer. $\endgroup$ Aug 2, 2013 at 7:38
  • $\begingroup$ Note that in your universal property, just like in any universal property, you should require that the group homomorphism $\psi$ be unique. This ensures that the group will be in some sense "minimal" and therefore uniquely determined (up to isomorphism) by the given semigroup. The point is that the direct product of your group $G$ with any other group would still satisfy your universal property. We want our group $G$ as general as possible, but also as small as possible; this corresponds to existence and uniqueness of $\psi$. $\endgroup$
    – Axel Boldt
    Dec 27, 2014 at 17:23

3 Answers 3

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What you are describing is the left adjoint of the forgetful functor from Group to Semigroup.

In the case of monoids and monoid homomorphisms, such a group is called the enveloping group of a monoid. You can find the description in George Bergman's Invitation to General Algebra and Universal Constructions, Chapter 3, Section 3.11, pages 65 and 66.

However, you don't always get embeddings.

For semigroups, the most natural thing is to adjoin a $1$, even if the semigroup already has one, and then perform the construction. If $S$ already had an identity, the construction will naturally collapse the new, adjoined, identity into the original one, and give you "the most general group into which you can map the semigroup".

Added. In fact, what the last construction is doing (for semigroups) is simply composing the two right adjoints: the right adjoint of the forgetful functor from Monoid to Semigroup is the functor that adjoins an identity (since even between monoids there are generally more semigroup homomorphisms than monoid homomorphisms). So if we look at the composition of the forgetful functors Group $\longrightarrow$ Monoid $\longrightarrow$ Semigroup, we obtain a right adjoint by composing the adjoints going the other way, Semigroup $\longrightarrow$ Monoid (adjoin a $1$), and Monoid $\longrightarrow$ Group (enveloping group). So: first adjoin a $1$, then construct the enveloping group.

(Interestingly, there is another functor from monoids to groups, namely the functor that assigns to every monoid its group of units, $M^*$. This functor is the right adjoint of the forgetful functor: given any monoid $M$ and any group $G$, there is a natural corespondence between $\mathbf{Monoid}(G,M)$ and $\mathbf{Group}(G,M^*)$).

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  • $\begingroup$ This becomes clearer in my mind. Thanks very much for your kind help. $\endgroup$ Nov 3, 2011 at 15:43
  • $\begingroup$ Is the composed forgetful functor $\bf Group \to \bf Semigroup$ full? faithful? Both? $\endgroup$ Jan 21, 2013 at 17:21
  • $\begingroup$ It seems to be that, in the Added note, all occurrences of "right adjoint" should be changed to "left adjoint". $\endgroup$
    – Uday Reddy
    May 8, 2013 at 11:26
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Update: I was initially wrong thinking that Grothendieck construction can be applied for any semigroup. Below a corrected version of my post.

If you start from a commutative monoid there is an easy way to construct a group, called Grothendieck construction. See http://en.wikipedia.org/wiki/Grothendieck_group. By the way, it's not always true that the monoid embeds into its Grothendieck group, but it suffices that it verifies the cancelation property: $a+b=a+c$ implies $b=c$.

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    $\begingroup$ Valerio, what you wrote about embedding of semigroups into groups is true only if the semigroup is commutative. The non-commutative case is more complicated, see e.g. Clifford, Preston: The Algebraic Theory of Semigroups - Page 36 or this paper. $\endgroup$ Oct 30, 2011 at 19:18
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    $\begingroup$ -1: The semigroup satisfying cancelation is not sufficient for the canonical map from the semigroup to the enveloping group (the unit of the left adjoin of the forgetful functor from groups to semigroups) to be an embedding. $\endgroup$ Oct 30, 2011 at 19:24
  • $\begingroup$ Yes, you're right! I was completely out. I've corrected the mistake, also if my answer remains probably not very pertinent. $\endgroup$ Oct 30, 2011 at 19:42
  • $\begingroup$ @Valerio: Just letting you know that I posted the following question: math.stackexchange.com/questions/79453/… I am not familiar with the construction of Grothendieck group, but a brief glance at the wikipedia article I got the feeling that it is related to that question. See also meta.math.stackexchange.com/questions/3159/… $\endgroup$ Nov 6, 2011 at 8:48
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Make a graph $a$ -> $aa$ for all elements. You will get a connected component around every idempotent, (where $a = aa$). Drop the tree nodes coming into the center of each connected component. What you have left is a group around each idempotent.

If you do this on the full transformation semigroup the largest connected component is the symmetric group; which has no trees to drop.

Visual Doing this on $T_{4}$. Upper left hand corner is $S_{4}$.

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  • $\begingroup$ Doesn't this just give you the maximal subgroup containing the given idempotent for each idempotent? $\endgroup$ Feb 1, 2017 at 17:44
  • $\begingroup$ Yes. From there you will have to slice and dice to the subgroups you want. $\endgroup$ Feb 1, 2017 at 17:58

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