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 § Formal definition.

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\neq 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 to the right.

Then the conditions:

$x^{2}=1\$

$x\neq 1\$

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

However the conditions:

$x^{4}=1\$

$x^{2}a^{-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 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:

$x_{1}^{2}g_{1}^{4}x_{3}=1$

$x_{3}^{2}g_{2}x_{4}g_{1}=1$

$\dots \$

The set $I\$ represents inequations like

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

$\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)\neq 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:

$g_{6}^{2}g_{1}^{4}g_{7}=1$

$g_{7}^{2}g_{2}g_{8}g_{1}=1$

$\dots \$

$g_{5}^{-1}g_{7}\neq 1$

$\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\$

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

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}\$.

See also

• Algebraic closure
  • Algebraically closed field

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)