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.

Informal discussion
Suppose we wished to find an element $$x\ $$ of a group $$G\ $$ satisfying the conditions (equations and inequations):


 * $$x^2=1\ $$
 * $$x^3=1\ $$
 * $$x\ne 1\ $$

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:


 * $$x^2=1\ $$
 * $$x\ne 1\ $$

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

However the conditions:


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

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:
 * $$x_1^2g_1^4x_3=1$$
 * $$x_3^2g_2x_4g_1=1$$
 * $$\dots\ $$

The set $$I\ $$ represents inequations like
 * $$g_5^{-1}x_3\ne 1$$
 * $$\dots\ $$

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 $$\tilde{f}(i)\ne 1\ $$ 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:
 * $$g_6^2g_1^4g_7=1$$
 * $$g_7^2g_2g_8g_1=1$$
 * $$\dots\ $$
 * $$g_5^{-1}g_7\ne 1$$
 * $$\dots\ $$

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\le j\le 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 $$C_{i+1}\ $$.