# When a semigroup extension $S \times A$ of a semigroup $S$ with a commutative group $A$ contains a copy of $S$?

Maybe is a trivial question, but I don't know how to handle it.

Setting: Let $S$ be a semigroup (i.e. has an associative operation with neutral element $e$) and let $(A,+)$ be a commutative group (with neutral element $0$).

A semigroup extension of $S$ with $A$ is any operation on $S \times A$, of the form

$(s,a)(s',a') = (s s' , a+a'+ \lambda(s,s')$

where $\lambda: S \times S \rightarrow A$ is a function such that $S \times A$ with the mentioned operation is a semigroup with neutral element $(e,0)$. Obviously, the operation is encoded by the function $\lambda$, which satisfies certain equations.

Problem: Describe the class of functions $\lambda$ with the property: there is an injective morphism of semigroups from $S$ to $S \times A$ (with the semigroup extension operation induced by $\lambda$). Is there an elegant way to describe this class?

Motivation: I am puzzled by the following two examples, I cannot put the finger on the essential difference between those.

Let $X$ and $Y$ be topological, locally convex, real vector spaces of dual variables $x \in X$ and $y \in Y$, with the duality product $\langle \cdot , \cdot \rangle : X \times Y \rightarrow \mathbb{R}$.

The spaces $X, Y$ have topologies compatible with the duality product, in the sense that for any continuous linear functional on $X$ there is an $y \in Y$ which puts the functional into the form $x \mapsto \langle x,y\rangle$ (respectively any continuous linear functional on $Y$ has the form $y \mapsto \langle x,y\rangle$, for a $x \in X$).

Example 1: (Heisenberg group) Let $S = X \times Y$ with the operation of addition of pairs of vectors and let $A = \mathbb{R}$ with addition. We may define a Heisenberg group over the pair $(X,Y)$ as $H(X,Y) = S \times A$ with the operation

$(x,y,a)(x',y',b) = (x+x', y+y', a+a'+ \langle x, y'\rangle - \langle x', y \rangle)$

There is no injective morphism from $(X\times Y, +)$ to $H(X,Y)$.

Example 2: Let this time $S = (X \times Y)^{*}$, the free semigroup generated by $X \times Y$, i.e. the collection of all finite words with letters from $X \times Y$, together with the empty word $e$, with the operation of concatenation of words.

Let $A$ be the set of bi-affine real functions on $X \times Y$, i.e. the collection of all functions $a: X \times Y \rightarrow \mathbb{R}$ which are affine and continuous in each argument. $A$ is a commutative group with the addition of real valued functions operation.

We define the function $\lambda: S \times S \rightarrow A$ by:

• $\lambda(e, c)(x,y) = \lambda(c,e)(x,y)=0$ for any $c \in S$ and any $(x,y) \in X \times Y$.
• if $c, h \in S$ are words $c = (x_{1},y_{1})...(x_{n}, y_{n})$ and $h = (u_{1},v_{1})...(u_{m}, v_{m})$, with $m,n \geq 1$, then

$\lambda(c,h)(x,y) = \langle u_{1} - x , y_{n} - y \rangle$ for any $(x,y) \in X \times Y$.

This $\lambda$ induces a semigroup extension operation on $S \times A$ and there is an injective morphism $F: S \rightarrow S \times A$, with the expression $F(c) = (c, E(c))$. Here, for any $c = (x_{1},y_{1})...(x_{n}, y_{n})$ the expression $E(c)(x,y)$ is the well known circular sum associated to the "discrete cycle" $(x_{1},y_{1})...(x_{n}, y_{n})(x,y)$, namely:

$E(c)(x,y) = \langle x_{1},y\rangle + \langle x,y_{n}\rangle - \langle x,y \rangle + \sum_{k=1}^{n-1}\langle x_{k+1}, y_{k}\rangle - \sum_{k=1}^{n} \langle x_{k},y_{k} \rangle$

dear to convex analysis.

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I am not great expert but terminology is confusing for me .... semidirect product always Contains copies of both groups see Wikipedia . You consier an extension. You might loook on p-complememt theorems on Wikipedia they give conditions for finite groups –  Alexander Chervov Oct 2 '12 at 15:26
Thanks for comment, I replaced "semidirect product" with "semigroup extension". –  Marius Buliga Oct 2 '12 at 18:16
General notation note: I've never come across "semigroup" requiring a neutral element. "Monoid" is what I've come across for "associative operation with neutral element", while semigroup is just "set with associative operation". –  Jonathan Kiehlmann Oct 3 '12 at 15:33
No, I've seen both names used for both objects, it's like $\mathbb{N}$, which contains or not $0$, according to pedagogical school of thought. –  Marius Buliga Oct 4 '12 at 7:04

First of all I would not call what you write a semidirect product. It will be such in your setting precisely when the natural maps to $S$ splits, in which case it will actually be isomorphic to a direct product.
More precisely, you are viewing $A$ here as a trivial module over $\mathbb ZS$. Then $\lambda$ is a 2-cocycle and you want it to be equivalent in $H^2(S,A)$ with the trivial cocycle.