Английская Википедия:Algebraically closed group
Шаблон:Short descriptionIn group theory, a group <math>A\ </math> is algebraically closed if any finite set of equations and inequations that are applicable to <math>A\ </math> have a solution in <math>A\ </math> without needing a group extension. This notion will be made precise later in the article in Шаблон:Slink.
Informal discussion
Suppose we wished to find an element <math>x\ </math> of a group <math>G\ </math> satisfying the conditions (equations and inequations):
- <math>x^2=1\ </math>
- <math>x^3=1\ </math>
- <math>x\ne 1\ </math>
Then it is easy to see that this is impossible because the first two equations imply <math>x=1\ </math>. In this case we say the set of conditions are inconsistent with <math>G\ </math>. (In fact this set of conditions are inconsistent with any group whatsoever.)
| <math>G\ </math> | |||||||||
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Now suppose <math>G\ </math> is the group with the multiplication table to the right.
Then the conditions:
- <math>x^2=1\ </math>
- <math>x\ne 1\ </math>
have a solution in <math>G\ </math>, namely <math>x=a\ </math>.
However the conditions:
- <math>x^4=1\ </math>
- <math>x^2a^{-1} = 1\ </math>
Do not have a solution in <math>G\ </math>, as can easily be checked.
| <math>H\ </math> | |||||||||||||||||||||||||
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However if we extend the group <math>G \ </math> to the group <math>H \ </math> with the adjacent multiplication table:
Then the conditions have two solutions, namely <math>x=b \ </math> and <math>x=c \ </math>.
Thus there are three possibilities regarding such conditions:
- They may be inconsistent with <math>G \ </math> and have no solution in any extension of <math>G \ </math>.
- They may have a solution in <math>G \ </math>.
- They may have no solution in <math>G \ </math> but nevertheless have a solution in some extension <math>H \ </math> of <math>G \ </math>.
It is reasonable to ask whether there are any groups <math>A \ </math> such that whenever a set of conditions like these have a solution at all, they have a solution in <math>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>G\ </math> is a group and <math>F\ </math> is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in <math>G\ </math> we mean a pair of subsets <math>E\ </math> and <math>I\ </math> of <math>F\star G</math> the free product of <math>F\ </math> and <math>G\ </math>.
This formalizes the notion of a set of equations and inequations consisting of variables <math>x_i\ </math> and elements <math>g_j\ </math> of <math>G\ </math>. The set <math>E\ </math> represents equations like:
- <math>x_1^2g_1^4x_3=1</math>
- <math>x_3^2g_2x_4g_1=1</math>
- <math>\dots\ </math>
The set <math>I\ </math> represents inequations like
- <math>g_5^{-1}x_3\ne 1</math>
- <math>\dots\ </math>
By a solution in <math>G\ </math> to this finite set of equations and inequations, we mean a homomorphism <math>f:F\rightarrow G</math>, such that <math>\tilde{f}(e)=1\ </math> for all <math>e\in E</math> and <math>\tilde{f}(i)\ne 1\ </math> for all <math>i\in I</math>, where <math>\tilde{f}</math> is the unique homomorphism <math>\tilde{f}:F\star G\rightarrow G</math> that equals <math>f\ </math> on <math>F\ </math> and is the identity on <math>G\ </math>.
This formalizes the idea of substituting elements of <math>G\ </math> for the variables to get true identities and inidentities. In the example the substitutions <math>x_1\mapsto g_6, x_3\mapsto g_7</math> and <math>x_4\mapsto g_8</math> yield:
- <math>g_6^2g_1^4g_7=1</math>
- <math>g_7^2g_2g_8g_1=1</math>
- <math>\dots\ </math>
- <math>g_5^{-1}g_7\ne 1</math>
- <math>\dots\ </math>
We say the finite set of equations and inequations is consistent with <math>G\ </math> if we can solve them in a "bigger" group <math>H\ </math>. More formally:
The equations and inequations are consistent with <math>G\ </math> if there is a group<math>H\ </math> and an embedding <math>h:G\rightarrow H</math> such that the finite set of equations and inequations <math>\tilde{h}(E)</math> and <math>\tilde{h}(I)</math> has a solution in <math>H\ </math>, where <math>\tilde{h}</math> is the unique homomorphism <math>\tilde{h}:F\star G\rightarrow F\star H</math> that equals <math>h\ </math> on <math>G\ </math> and is the identity on <math>F\ </math>.
Now we formally define the group <math>A\ </math> to be algebraically closed if every finite set of equations and inequations that has coefficients in <math>A\ </math> and is consistent with <math>A\ </math> has a solution in <math>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>C\ </math> can be embedded in an algebraically closed group follows.
First we embed <math>C\ </math> in a countable group <math>C_1\ </math> with the property that every finite set of equations with coefficients in <math>C\ </math> that is consistent in <math>C_1\ </math> has a solution in <math>C_1\ </math> as follows:
There are only countably many finite sets of equations and inequations with coefficients in <math>C\ </math>. Fix an enumeration <math>S_0,S_1,S_2,\dots\ </math> of them. Define groups <math>D_0,D_1,D_2,\dots\ </math> inductively by:
- <math>D_0 = C\ </math>
- <math>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>C_1=\cup_{i=0}^{\infty}D_{i}</math>
Now iterate this construction to get a sequence of groups <math>C=C_0,C_1,C_2,\dots\ </math> and let:
- <math>A=\cup_{i=0}^{\infty}C_{i}</math>
Then <math>A\ </math> is a countable group containing <math>C\ </math>. It is algebraically closed because any finite set of equations and inequations that is consistent with <math>A\ </math> must have coefficients in some <math>C_i\ </math> and so must have a solution in <math>C_{i+1}\ </math>.
See also
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)
