Algebraically closed group

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Short description: Group allowing solution of all algebraic equations

In group theory, a group [math]\displaystyle{ A\ }[/math] is algebraically closed if any finite set of equations and inequations that are applicable to [math]\displaystyle{ A\ }[/math] have a solution in [math]\displaystyle{ A\ }[/math] 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 [math]\displaystyle{ x\ }[/math] of a group [math]\displaystyle{ G\ }[/math] satisfying the conditions (equations and inequations):

[math]\displaystyle{ x^2=1\ }[/math]
[math]\displaystyle{ x^3=1\ }[/math]
[math]\displaystyle{ x\ne 1\ }[/math]

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

[math]\displaystyle{ G\ }[/math]
[math]\displaystyle{ . \ }[/math] [math]\displaystyle{ \underline{1} \ }[/math] [math]\displaystyle{ \underline{a} \ }[/math]
[math]\displaystyle{ \underline{1} \ }[/math] [math]\displaystyle{ 1 \ }[/math] [math]\displaystyle{ a \ }[/math]
[math]\displaystyle{ \underline{a} \ }[/math] [math]\displaystyle{ a \ }[/math] [math]\displaystyle{ 1 \ }[/math]

Now suppose [math]\displaystyle{ G\ }[/math] is the group with the multiplication table to the right.

Then the conditions:

[math]\displaystyle{ x^2=1\ }[/math]
[math]\displaystyle{ x\ne 1\ }[/math]

have a solution in [math]\displaystyle{ G\ }[/math], namely [math]\displaystyle{ x=a\ }[/math].

However the conditions:

[math]\displaystyle{ x^4=1\ }[/math]
[math]\displaystyle{ x^2a^{-1} = 1\ }[/math]

Do not have a solution in [math]\displaystyle{ G\ }[/math], as can easily be checked.

[math]\displaystyle{ H\ }[/math]
[math]\displaystyle{ . \ }[/math] [math]\displaystyle{ \underline{1} \ }[/math] [math]\displaystyle{ \underline{a} \ }[/math] [math]\displaystyle{ \underline{b} \ }[/math] [math]\displaystyle{ \underline{c} \ }[/math]
[math]\displaystyle{ \underline{1} \ }[/math] [math]\displaystyle{ 1 \ }[/math] [math]\displaystyle{ a \ }[/math] [math]\displaystyle{ b \ }[/math] [math]\displaystyle{ c \ }[/math]
[math]\displaystyle{ \underline{a} \ }[/math] [math]\displaystyle{ a \ }[/math] [math]\displaystyle{ 1 \ }[/math] [math]\displaystyle{ c \ }[/math] [math]\displaystyle{ b \ }[/math]
[math]\displaystyle{ \underline{b} \ }[/math] [math]\displaystyle{ b \ }[/math] [math]\displaystyle{ c \ }[/math] [math]\displaystyle{ a \ }[/math] [math]\displaystyle{ 1 \ }[/math]
[math]\displaystyle{ \underline{c} \ }[/math] [math]\displaystyle{ c \ }[/math] [math]\displaystyle{ b \ }[/math] [math]\displaystyle{ 1 \ }[/math] [math]\displaystyle{ a \ }[/math]

However if we extend the group [math]\displaystyle{ G \ }[/math] to the group [math]\displaystyle{ H \ }[/math] with the adjacent multiplication table:

Then the conditions have two solutions, namely [math]\displaystyle{ x=b \ }[/math] and [math]\displaystyle{ x=c \ }[/math].

Thus there are three possibilities regarding such conditions:

  • They may be inconsistent with [math]\displaystyle{ G \ }[/math] and have no solution in any extension of [math]\displaystyle{ G \ }[/math].
  • They may have a solution in [math]\displaystyle{ G \ }[/math].
  • They may have no solution in [math]\displaystyle{ G \ }[/math] but nevertheless have a solution in some extension [math]\displaystyle{ H \ }[/math] of [math]\displaystyle{ G \ }[/math].

It is reasonable to ask whether there are any groups [math]\displaystyle{ A \ }[/math] such that whenever a set of conditions like these have a solution at all, they have a solution in [math]\displaystyle{ A \ }[/math] 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 [math]\displaystyle{ G\ }[/math] is a group and [math]\displaystyle{ F\ }[/math] is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in [math]\displaystyle{ G\ }[/math] we mean a pair of subsets [math]\displaystyle{ E\ }[/math] and [math]\displaystyle{ I\ }[/math] of [math]\displaystyle{ F\star G }[/math] the free product of [math]\displaystyle{ F\ }[/math] and [math]\displaystyle{ G\ }[/math].

This formalizes the notion of a set of equations and inequations consisting of variables [math]\displaystyle{ x_i\ }[/math] and elements [math]\displaystyle{ g_j\ }[/math] of [math]\displaystyle{ G\ }[/math]. The set [math]\displaystyle{ E\ }[/math] represents equations like:

[math]\displaystyle{ x_1^2g_1^4x_3=1 }[/math]
[math]\displaystyle{ x_3^2g_2x_4g_1=1 }[/math]
[math]\displaystyle{ \dots\ }[/math]

The set [math]\displaystyle{ I\ }[/math] represents inequations like

[math]\displaystyle{ g_5^{-1}x_3\ne 1 }[/math]
[math]\displaystyle{ \dots\ }[/math]

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

This formalizes the idea of substituting elements of [math]\displaystyle{ G\ }[/math] for the variables to get true identities and inidentities. In the example the substitutions [math]\displaystyle{ x_1\mapsto g_6, x_3\mapsto g_7 }[/math] and [math]\displaystyle{ x_4\mapsto g_8 }[/math] yield:

[math]\displaystyle{ g_6^2g_1^4g_7=1 }[/math]
[math]\displaystyle{ g_7^2g_2g_8g_1=1 }[/math]
[math]\displaystyle{ \dots\ }[/math]
[math]\displaystyle{ g_5^{-1}g_7\ne 1 }[/math]
[math]\displaystyle{ \dots\ }[/math]

We say the finite set of equations and inequations is consistent with [math]\displaystyle{ G\ }[/math] if we can solve them in a "bigger" group [math]\displaystyle{ H\ }[/math]. More formally:

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

Now we formally define the group [math]\displaystyle{ A\ }[/math] to be algebraically closed if every finite set of equations and inequations that has coefficients in [math]\displaystyle{ A\ }[/math] and is consistent with [math]\displaystyle{ A\ }[/math] has a solution in [math]\displaystyle{ A\ }[/math].

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 [math]\displaystyle{ C\ }[/math] can be embedded in an algebraically closed group follows.

First we embed [math]\displaystyle{ C\ }[/math] in a countable group [math]\displaystyle{ C_1\ }[/math] with the property that every finite set of equations with coefficients in [math]\displaystyle{ C\ }[/math] that is consistent in [math]\displaystyle{ C_1\ }[/math] has a solution in [math]\displaystyle{ C_1\ }[/math] as follows:

There are only countably many finite sets of equations and inequations with coefficients in [math]\displaystyle{ C\ }[/math]. Fix an enumeration [math]\displaystyle{ S_0,S_1,S_2,\dots\ }[/math] of them. Define groups [math]\displaystyle{ D_0,D_1,D_2,\dots\ }[/math] inductively by:

[math]\displaystyle{ D_0 = C\ }[/math]
[math]\displaystyle{ 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. }[/math]

Now let:

[math]\displaystyle{ C_1=\cup_{i=0}^{\infty}D_{i} }[/math]

Now iterate this construction to get a sequence of groups [math]\displaystyle{ C=C_0,C_1,C_2,\dots\ }[/math] and let:

[math]\displaystyle{ A=\cup_{i=0}^{\infty}C_{i} }[/math]

Then [math]\displaystyle{ A\ }[/math] is a countable group containing [math]\displaystyle{ C\ }[/math]. It is algebraically closed because any finite set of equations and inequations that is consistent with [math]\displaystyle{ A\ }[/math] must have coefficients in some [math]\displaystyle{ C_i\ }[/math] and so must have a solution in [math]\displaystyle{ C_{i+1}\ }[/math].

See also


  • 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)