# Algebraically closed group

Short description: Group allowing solution of all algebraic equations

In group theory, a group $\displaystyle{ A\ }$ is algebraically closed if any finite set of equations and inequations that are applicable to $\displaystyle{ A\ }$ 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):

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

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 to the right.

Then the conditions:

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

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

However the conditions:

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

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 the adjacent 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:

$\displaystyle{ x_1^2g_1^4x_3=1 }$
$\displaystyle{ x_3^2g_2x_4g_1=1 }$
$\displaystyle{ \dots\ }$

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

$\displaystyle{ g_5^{-1}x_3\ne 1 }$
$\displaystyle{ \dots\ }$

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 $\displaystyle{ \tilde{f}(i)\ne 1\ }$ 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:

$\displaystyle{ g_6^2g_1^4g_7=1 }$
$\displaystyle{ g_7^2g_2g_8g_1=1 }$
$\displaystyle{ \dots\ }$
$\displaystyle{ g_5^{-1}g_7\ne 1 }$
$\displaystyle{ \dots\ }$

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:

• 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 $\displaystyle{ C\ }$ 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:

$\displaystyle{ D_0 = C\ }$
$\displaystyle{ D_{i+1} = \left\{\begin{matrix} D_i\ &\mbox{if}\ S_i\ \mbox{is not consistent with}\ D_i \\ \langle D_i,h_1,h_2,\dots,h_n \rangle &\mbox{if}\ S_i\ \mbox{has a solution in}\ H\supseteq D_i\ \mbox{with}\ x_j\mapsto h_j\ 1\le j\le n \end{matrix}\right. }$

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 $\displaystyle{ C_{i+1}\ }$.