Algebraically closed group

From Wikipedia, the free encyclopedia

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.

Contents

[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 \\ <D_i,h_1,h_2,\dots,h_n> &\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)
Retrieved from "http://en.wikipedia.org../../../a/l/g/Algebraically_closed_group.html"