Algebraically closed group

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In mathematics, in the realm of group theory, a group $A\$ is algebraically closed if any finite set of equations and inequations that "make sense" in $A\$ already have a solution in $A\$ . This idea will be made precise later in the article.

1 Informal discussion
2 Formal definition of an algebraically closed group
3 Known Results
4 References

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.)

$G\$

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

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

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}\ne 1\$

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

$H\$

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

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

Then the condions have two solutions, namely $x=b \$ and $x=c \$ .

Thus there are three possibilities regarding such conditions:

The 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.

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\$ .

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 solvable word problem if and only if it can embedded in every algebraically closed group.

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

First we embedd $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 countable 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\$

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

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}\$ .

A. Macintyre: On algebraically closed groups, ann. of Math, 96, 53-97 (1972)
B.H. Neumann: A note on algebraically closed groups. J. London Math. Soc. 27, 227-242 (1952)
B.H. Neumann: The isomorphism problem for algebraically closed groups. In: Word Problems, pp 553-562. Amsterdam: North-Holland 1973
W.R. Scott: Algebraically closed groups. Proc. Amer. Math. Soc. 2, 118-121 (1951)

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