# Algebraically closed group

In group theory, a group $A\$ is algebraically closed if any finite set of equations and inequations that are applicable to $A\$ have a solution in $A\$ without needing a group extension. This notion will be made precise later in the article in § Formal definition.

## Informal discussion

Suppose we wished to find an element $x\$ of a group $G\$ satisfying the conditions (equations and inequations):
       $x^{2}=1\$
       $x^{3}=1\$
       $x\neq 1\$
Then it is easy to see that this is impossible because the first two equations imply $x=1\$. In this case we say the set of conditions are inconsistent with $G\$. (In fact this set of conditions are inconsistent with any group whatsoever.)
Now suppose $G\$ is the group with the multiplication table to the right.
Then the conditions:
       $x^{2}=1\$
       $x\neq 1\$
have a solution in $G\$, namely $x=a\$.
However the conditions:
       $x^{4}=1\$
       $x^{2}a^{-1}=1\$
Do not have a solution in $G\$, as can easily be checked.
However, if we extend the group $G\$ to the group $H\$ with the adjacent multiplication table:
Then the conditions have two solutions, namely $x=b\$ and $x=c\$.
Thus there are three possibilities regarding such conditions:
* They may be inconsistent with $G\$ and have no solution in any extension of $G\$.
* They may have a solution in $G\$.
* They may have no solution in $G\$ but nevertheless have a solution in some extension $H\$ of $G\$.
It is reasonable to ask whether there are any groups $A\$ such that whenever a set of conditions like these have a solution at all, they have a solution in $A\$ itself? The answer turns out to be "yes", and we call such groups algebraically closed groups.

## Formal definition

We first need some preliminary ideas.
If $G\$ is a group and $F\$ is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in $G\$ we mean a pair of subsets $E\$ and $I\$ of $F\star G$ the free product of $F\$ and $G\$.
This formalizes the notion of a set of equations and inequations consisting of variables $x_{i}\$ and elements $g_{j}\$ of $G\$. The set $E\$ represents equations like:
       $x_{1}^{2}g_{1}^{4}x_{3}=1$
       $x_{3}^{2}g_{2}x_{4}g_{1}=1$
       $\dots \$
The set $I\$ represents inequations like
       $g_{5}^{-1}x_{3}\neq 1$
       $\dots \$
By a solution in $G\$ to this finite set of equations and inequations, we mean a homomorphism $f:F\rightarrow G$, such that ${\tilde {f}}(e)=1\$ for all $e\in E$ and ${\tilde {f}}(i)\neq 1\$ for all $i\in I$, where ${\tilde {f}}$ is the unique homomorphism ${\tilde {f}}:F\star G\rightarrow G$ that equals $f\$ on $F\$ and is the identity on $G\$.
This formalizes the idea of substituting elements of $G\$ for the variables to get true identities and inidentities. In the example the substitutions $x_{1}\mapsto g_{6},x_{3}\mapsto g_{7}$ and $x_{4}\mapsto g_{8}$ yield:
       $g_{6}^{2}g_{1}^{4}g_{7}=1$
       $g_{7}^{2}g_{2}g_{8}g_{1}=1$
       $\dots \$
       $g_{5}^{-1}g_{7}\neq 1$
       $\dots \$
We say the finite set of equations and inequations is consistent with $G\$ if we can solve them in a "bigger" group $H\$. More formally:
The equations and inequations are consistent with $G\$ if there is a group$H\$ and an embedding $h:G\rightarrow H$ such that the finite set of equations and inequations ${\tilde {h}}(E)$ and ${\tilde {h}}(I)$ has a solution in $H\$, where ${\tilde {h}}$ is the unique homomorphism ${\tilde {h}}:F\star G\rightarrow F\star H$ that equals $h\$ on $G\$ and is the identity on $F\$.
Now we formally define the group $A\$ to be algebraically closed if every finite set of equations and inequations that has coefficients in $A\$ and is consistent with $A\$ has a solution in $A\$.

## Known results

It is difficult to give concrete examples of algebraically closed groups as the following results indicate:
* Every countable group can be embedded in a countable algebraically closed group.
* Every algebraically closed group is simple.
* No algebraically closed group is finitely generated.
* An algebraically closed group cannot be recursively presented.
* A finitely generated group has a solvable word problem if and only if it can be embedded in every algebraically closed group.
The proofs of these results are in general very complex. However, a sketch of the proof that a countable group $C\$ can be embedded in an algebraically closed group follows.
First we embed $C\$ in a countable group $C_{1}\$ with the property that every finite set of equations with coefficients in $C\$ that is consistent in $C_{1}\$ has a solution in $C_{1}\$ as follows:
There are only countably many finite sets of equations and inequations with coefficients in $C\$. Fix an enumeration $S_{0},S_{1},S_{2},\dots \$ of them. Define groups $D_{0},D_{1},D_{2},\dots \$ inductively by:
       $D_{0}=C\$
       $D_{i+1}=\left\{{\begin{matrix}D_{i}\ &{\mbox{if}}\ S_{i}\ {\mbox{is not consistent with}}\ D_{i}\\\langle D_{i},h_{1},h_{2},\dots ,h_{n}\rangle &{\mbox{if}}\ S_{i}\ {\mbox{has a solution in}}\ H\supseteq D_{i}\ {\mbox{with}}\ x_{j}\mapsto h_{j}\ 1\leq j\leq n\end{matrix}}\right.$
Now let:
       $C_{1}=\cup _{i=0}^{\infty }D_{i}$
Now iterate this construction to get a sequence of groups $C=C_{0},C_{1},C_{2},\dots \$ and let:
       $A=\cup _{i=0}^{\infty }C_{i}$
Then $A\$ is a countable group containing $C\$. It is algebraically closed because any finite set of equations and inequations that is consistent with $A\$ must have coefficients in some $C_{i}\$ and so must have a solution in $C_{i+1}\$.