Algebraically closed group

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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.

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[edit] 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}\ne 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 condions have two solutions, namely x=b \ and x=c \ .

Thus there are three possibilities regarding such conditions:

  • The 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.

[edit] 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 groupH\ 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\ .

[edit] Known Results

It is difficult to give concrete examples of algebraically closed groups as the following results indicate:

The proofs of these results are, in general very complex. However a sketch the proof that a countable group C\ can be embedded in an algebraically closed group follows.

First we embedd 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 countable 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}\ .

[edit] 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)