algebraically closed group

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

::$x^2=1\backslash$

::$x^3=1\backslash$

::$x\backslash ne\; 1\backslash$

Then it is easy to see that this is impossible because the first two equations imply $x=1\backslash$. In this case we say the set of conditions are inconsistent with $G\backslash$. (In fact this set of conditions are inconsistent with any group whatsoever.)

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Now suppose $G\backslash$ is the group with the multiplication table to the right.

Then the conditions:

::$x^2=1\backslash$

::$x\backslash ne\; 1\backslash$

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

However the conditions:

::$x^4=1\backslash$

::$x^2a^\{-1\}\; =\; 1\backslash$

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

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However, if we extend the group $G\; \backslash$ to the group $H\; \backslash$ with the adjacent multiplication table:

Then the conditions have two solutions, namely $x=b\; \backslash$ and $x=c\; \backslash$.

Thus there are three possibilities regarding such conditions:

• They may be inconsistent with $G\; \backslash$ and have no solution in any extension of $G\; \backslash$.
• They may have a solution in $G\; \backslash$.
• They may have no solution in $G\; \backslash$ but nevertheless have a solution in some extension $H\; \backslash$ of $G\; \backslash$.

It is reasonable to ask whether there are any groups $A\; \backslash$ such that whenever a set of conditions like these have a solution at all, they have a solution in $A\; \backslash$ itself? The answer turns out to be "yes", and we call such groups algebraically closed groups.

We first need some preliminary ideas.

If $G\backslash$ is a group and $F\backslash$ is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in $G\backslash$ we mean a pair of subsets $E\backslash$ and $I\backslash$ of $F\backslash star\; G$ the free product of $F\backslash$ and $G\backslash$.

This formalizes the notion of a set of equations and inequations consisting of variables $x\_i\backslash$ and elements $g\_j\backslash$ of $G\backslash$. The set $E\backslash$ represents equations like:

::$x\_1^2g\_1^4x\_3=1$

::$x\_3^2g\_2x\_4g\_1=1$

::$\backslash dots\backslash$

The set $I\backslash$ represents inequations like

::$g\_5^\{-1\}x\_3\backslash ne\; 1$

::$\backslash dots\backslash$

By a solution in $G\backslash$ to this finite set of equations and inequations, we mean a homomorphism $f:F\backslash rightarrow\; G$, such that $\backslash tilde\{f\}(e)=1\backslash$ for all $e\backslash in\; E$ and $\backslash tilde\{f\}(i)\backslash ne\; 1\backslash$ for all $i\backslash in\; I$, where $\backslash tilde\{f\}$ is the unique homomorphism $\backslash tilde\{f\}:F\backslash star\; G\backslash rightarrow\; G$ that equals $f\backslash$ on $F\backslash$ and is the identity on $G\backslash$.

This formalizes the idea of substituting elements of $G\backslash$ for the variables to get true identities and inidentities. In the example the substitutions $x\_1\backslash mapsto\; g\_6,\; x\_3\backslash mapsto\; g\_7$ and $x\_4\backslash mapsto\; g\_8$ yield:

::$g\_6^2g\_1^4g\_7=1$

::$g\_7^2g\_2g\_8g\_1=1$

::$\backslash dots\backslash$

::$g\_5^\{-1\}g\_7\backslash ne\; 1$

::$\backslash dots\backslash$

We say the finite set of equations and inequations is consistent with $G\backslash$ if we can solve them in a "bigger" group $H\backslash$. More formally:

The equations and inequations are consistent with $G\backslash$ if there is a group$H\backslash$ and an embedding $h:G\backslash rightarrow\; H$ such that the finite set of equations and inequations $\backslash tilde\{h\}(E)$ and $\backslash tilde\{h\}(I)$ has a solution in $H\backslash$, where $\backslash tilde\{h\}$ is the unique homomorphism $\backslash tilde\{h\}:F\backslash star\; G\backslash rightarrow\; F\backslash star\; H$ that equals $h\backslash$ on $G\backslash$ and is the identity on $F\backslash$.

Now we formally define the group $A\backslash$ to be algebraically closed if every finite set of equations and inequations that has coefficients in $A\backslash$ and is consistent with $A\backslash$ has a solution in $A\backslash$.

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 $C\backslash$ can be embedded in an algebraically closed group follows.

First we embed $C\backslash$ in a countable group $C\_1\backslash$ with the property that every finite set of equations with coefficients in $C\backslash$ that is consistent in $C\_1\backslash$ has a solution in $C\_1\backslash$ as follows:

There are only countably many finite sets of equations and inequations with coefficients in $C\backslash$. Fix an enumeration $S\_0,S\_1,S\_2,\backslash dots\backslash$ of them. Define groups $D\_0,D\_1,D\_2,\backslash dots\backslash$ inductively by:

::$D\_0\; =\; C\backslash$

::$D\_\{i+1\}\; =$

\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:

::$C\_1=\backslash cup\_\{i=0\}^\{\backslash infty\}D\_\{i\}$

Now iterate this construction to get a sequence of groups $C=C\_0,C\_1,C\_2,\backslash dots\backslash$ and let:

::$A=\backslash cup\_\{i=0\}^\{\backslash infty\}C\_\{i\}$

Then $A\backslash$ is a countable group containing $C\backslash$. It is algebraically closed because any finite set of equations and inequations that is consistent with $A\backslash$ must have coefficients in some $C\_i\backslash$ and so must have a solution in $C\_\{i+1\}\backslash$.

• Algebraic closure
• Algebraically closed field

• 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)

Category:Properties of groups