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):

$x^2=1$

$x^3=1$

$x\ne 1$

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.)

$G$
$.$ $\underset{\_}{1}$ $\underset{\_}{a}$ $\underset{\_}{1}$ $1$ $a$ $\underset{\_}{a}$ $a$ $1$

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:

$x^4=1$

$x^2{a}^{-1}=1$

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

$H$
$.$ $\underset{\_}{1}$ $\underset{\_}{a}$ $\underset{\_}{b}$ $\underset{\_}{c}$ $\underset{\_}{1}$ $1$ $a$ $b$ $c$ $\underset{\_}{a}$ $a$ $1$ $c$ $b$ $\underset{\_}{b}$ $b$ $c$ $a$ $1$ $\underset{\_}{c}$ $c$ $b$ $1$ $a$

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

$g_5^{-1}{x}_3\ne 1$

$\dots$

By a solution in $G$ to this finite set of equations and inequations, we mean a homomorphism $f:F\to G$, 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 $\tilde{f}:F\star G\to G$ 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:

$g_6^2{g}_1^4{g}_7=1$

$g_7^2{g}_2{g}_8{g}_1=1$

$\dots$

$g_5^{-1}{g}_7\ne 1$

$\dots$

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 $h:G\to H$ 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 $\tilde{h}:F\star G\to F\star H$ 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:

$D_0=C$

$$\begin{gathered} D_{i+1}=\{\begin{array}{cc}{D}_i \\ & \text{if} \\ {S}_i \\ \text{is not consistent with} \\ {D}_i\\ ⟨D_i,h_1,h_2,\dots ,h_n⟩& \text{if} \\ {S}_i \\ \text{has a solution in} \\ H\supseteq D_i \\ \text{with} \\ {x}_j\mapsto h_j \\ 1\le j\le n\end{array} \end{gathered}$$

Now let:

$C_1=\underset{i=0}{\overset{\mathrm{\infty }}{\cup }}{D}_i$

Now iterate this construction to get a sequence of groups $C=C_0,C_1,C_2,\dots$ and let:

$A=\underset{i=0}{\overset{\mathrm{\infty }}{\cup }}{C}_i$

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)