# Algebraically closed group

In group theory, a group ${\displaystyle A\ }$ is algebraically closed if any finite set of equations and inequations that "make sense" in ${\displaystyle A\ }$ have a solution in ${\displaystyle A\ }$ 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 ${\displaystyle x\ }$ of a group ${\displaystyle G\ }$ satisfying the conditions (equations and inequations):

${\displaystyle x^{2}=1\ }$
${\displaystyle x^{3}=1\ }$
${\displaystyle x\neq 1\ }$

Then it is easy to see that this is impossible because the first two equations imply ${\displaystyle x=1\ }$. In this case we say the set of conditions are inconsistent with ${\displaystyle G\ }$. (In fact this set of conditions are inconsistent with any group whatsoever.)

${\displaystyle G\ }$
${\displaystyle {\underline {1}}\ }$ ${\displaystyle {\underline {a}}\ }$ ${\displaystyle .\ }$ ${\displaystyle 1\ }$ ${\displaystyle a\ }$ ${\displaystyle a\ }$ ${\displaystyle 1\ }$

Now suppose ${\displaystyle G\ }$ is the group with the multiplication table:

Then the conditions:

${\displaystyle x^{2}=1\ }$
${\displaystyle x\neq 1\ }$

have a solution in ${\displaystyle G\ }$, namely ${\displaystyle x=a\ }$.

However the conditions:

${\displaystyle x^{4}=1\ }$
${\displaystyle x^{2}a^{-1}=1\ }$

Do not have a solution in ${\displaystyle G\ }$, as can easily be checked.

${\displaystyle H\ }$
${\displaystyle {\underline {1}}\ }$ ${\displaystyle {\underline {a}}\ }$ ${\displaystyle {\underline {b}}\ }$ ${\displaystyle {\underline {c}}\ }$ ${\displaystyle .\ }$ ${\displaystyle 1\ }$ ${\displaystyle a\ }$ ${\displaystyle b\ }$ ${\displaystyle c\ }$ ${\displaystyle a\ }$ ${\displaystyle 1\ }$ ${\displaystyle c\ }$ ${\displaystyle b\ }$ ${\displaystyle b\ }$ ${\displaystyle c\ }$ ${\displaystyle a\ }$ ${\displaystyle 1\ }$ ${\displaystyle c\ }$ ${\displaystyle b\ }$ ${\displaystyle 1\ }$ ${\displaystyle a\ }$

However if we extend the group ${\displaystyle G\ }$ to the group ${\displaystyle H\ }$ with multiplication table:

Then the conditions have two solutions, namely ${\displaystyle x=b\ }$ and ${\displaystyle x=c\ }$.

Thus there are three possibilities regarding such conditions:

• They may be inconsistent with ${\displaystyle G\ }$ and have no solution in any extension of ${\displaystyle G\ }$.
• They may have a solution in ${\displaystyle G\ }$.
• They may have no solution in ${\displaystyle G\ }$ but nevertheless have a solution in some extension ${\displaystyle H\ }$ of ${\displaystyle G\ }$.

It is reasonable to ask whether there are any groups ${\displaystyle A\ }$ such that whenever a set of conditions like these have a solution at all, they have a solution in ${\displaystyle A\ }$ 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 ${\displaystyle G\ }$ is a group and ${\displaystyle F\ }$ is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in ${\displaystyle G\ }$ we mean a pair of subsets ${\displaystyle E\ }$ and ${\displaystyle I\ }$ of ${\displaystyle F\star G}$ the free product of ${\displaystyle F\ }$ and ${\displaystyle G\ }$.

This formalizes the notion of a set of equations and inequations consisting of variables ${\displaystyle x_{i}\ }$ and elements ${\displaystyle g_{j}\ }$ of ${\displaystyle G\ }$. The set ${\displaystyle E\ }$ represents equations like:

${\displaystyle x_{1}^{2}g_{1}^{4}x_{3}=1}$
${\displaystyle x_{3}^{2}g_{2}x_{4}g_{1}=1}$
${\displaystyle \dots \ }$

The set ${\displaystyle I\ }$ represents inequations like

${\displaystyle g_{5}^{-1}x_{3}\neq 1}$
${\displaystyle \dots \ }$

By a solution in ${\displaystyle G\ }$ to this finite set of equations and inequations, we mean a homomorphism ${\displaystyle f:F\rightarrow G}$, such that ${\displaystyle {\tilde {f}}(e)=1\ }$ for all ${\displaystyle e\in E}$ and ${\displaystyle {\tilde {f}}(i)\neq 1\ }$ for all ${\displaystyle i\in I}$, where ${\displaystyle {\tilde {f}}}$ is the unique homomorphism ${\displaystyle {\tilde {f}}:F\star G\rightarrow G}$ that equals ${\displaystyle f\ }$ on ${\displaystyle F\ }$ and is the identity on ${\displaystyle G\ }$.

This formalizes the idea of substituting elements of ${\displaystyle G\ }$ for the variables to get true identities and inidentities. In the example the substitutions ${\displaystyle x_{1}\mapsto g_{6},x_{3}\mapsto g_{7}}$ and ${\displaystyle x_{4}\mapsto g_{8}}$ yield:

${\displaystyle g_{6}^{2}g_{1}^{4}g_{7}=1}$
${\displaystyle g_{7}^{2}g_{2}g_{8}g_{1}=1}$
${\displaystyle \dots \ }$
${\displaystyle g_{5}^{-1}g_{7}\neq 1}$
${\displaystyle \dots \ }$

We say the finite set of equations and inequations is consistent with ${\displaystyle G\ }$ if we can solve them in a "bigger" group ${\displaystyle H\ }$. More formally:

The equations and inequations are consistent with ${\displaystyle G\ }$ if there is a group${\displaystyle H\ }$ and an embedding ${\displaystyle h:G\rightarrow H}$ such that the finite set of equations and inequations ${\displaystyle {\tilde {h}}(E)}$ and ${\displaystyle {\tilde {h}}(I)}$ has a solution in ${\displaystyle H\ }$, where ${\displaystyle {\tilde {h}}}$ is the unique homomorphism ${\displaystyle {\tilde {h}}:F\star G\rightarrow F\star H}$ that equals ${\displaystyle h\ }$ on ${\displaystyle G\ }$ and is the identity on ${\displaystyle F\ }$.

Now we formally define the group ${\displaystyle A\ }$ to be algebraically closed if every finite set of equations and inequations that has coefficients in ${\displaystyle A\ }$ and is consistent with ${\displaystyle A\ }$ has a solution in ${\displaystyle A\ }$.

## Known Results

It is difficult to give concrete examples of algebraically closed groups as the following results indicate:

The proofs of these results are in general very complex. However, a sketch of the proof that a countable group ${\displaystyle C\ }$ can be embedded in an algebraically closed group follows.

First we embed ${\displaystyle C\ }$ in a countable group ${\displaystyle C_{1}\ }$ with the property that every finite set of equations with coefficients in ${\displaystyle C\ }$ that is consistent in ${\displaystyle C_{1}\ }$ has a solution in ${\displaystyle C_{1}\ }$ as follows:

There are only countably many finite sets of equations and inequations with coefficients in ${\displaystyle C\ }$. Fix an enumeration ${\displaystyle S_{0},S_{1},S_{2},\dots \ }$ of them. Define groups ${\displaystyle D_{0},D_{1},D_{2},\dots \ }$ inductively by:

${\displaystyle D_{0}=C\ }$
${\displaystyle D_{i+1}=\left\{{\begin{matrix}D_{i}\ &{\mbox{if}}\ S_{i}\ {\mbox{is not consistent with}}\ D_{i}\\\langle D_{i},h_{1},h_{2},\dots ,h_{n}\rangle &{\mbox{if}}\ S_{i}\ {\mbox{has a solution in}}\ H\supseteq D_{i}\ {\mbox{with}}\ x_{j}\mapsto h_{j}\ 1\leq j\leq n\end{matrix}}\right.}$

Now let:

${\displaystyle C_{1}=\cup _{i=0}^{\infty }D_{i}}$

Now iterate this construction to get a sequence of groups ${\displaystyle C=C_{0},C_{1},C_{2},\dots \ }$ and let:

${\displaystyle A=\cup _{i=0}^{\infty }C_{i}}$

Then ${\displaystyle A\ }$ is a countable group containing ${\displaystyle C\ }$. It is algebraically closed because any finite set of equations and inequations that is consistent with ${\displaystyle A\ }$ must have coefficients in some ${\displaystyle C_{i}\ }$ and so must have a solution in ${\displaystyle C_{i+1}\ }$.