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Algebraically closed group

Group allowing solution of all algebraic equations


Group allowing solution of all algebraic equations

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 .

Informal discussion

Suppose we wished to find an element x\ of a group G\ satisfying the conditions (equations and inequations):

::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.) ::data[format=table] | \underline{1} \ | \underline{a} \ | \underline{1} \ | \underline{a} \ | |---|---|---|---| | . \ | | | | | 1 \ | 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^2a^{-1} = 1\ Do not have a solution in G\ , as can easily be checked. ::data[format=table] | \underline{1} \ | \underline{a} \ | \underline{b} \ | \underline{c} \ | \underline{1} \ | \underline{a} \ | \underline{b} \ | \underline{c} \ | |---|---|---|---|---|---|---|---| | . \ | | | | | | | | | 1 \ | a \ | b \ | c \ | | | | | | a \ | 1 \ | c \ | b \ | | | | | | b \ | c \ | a \ | 1 \ | | | | | | 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_1^2g_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\ . ## 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\ ::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}\ . ## 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) ::callout[type=info title="Wikipedia Source"] This article was imported from [Wikipedia](https://en.wikipedia.org/wiki/Algebraically_closed_group) and is available under the [Creative Commons Attribution-ShareAlike 4.0 License](https://creativecommons.org/licenses/by-sa/4.0/). Content has been adapted to SurfDoc format. Original contributors can be found on the [article history page](https://en.wikipedia.org/wiki/Algebraically_closed_group?action=history). ::
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