Algebraically closed group

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.

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.)

$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} = 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 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 of an algebraically closed group

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}\$.

References

• 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)