Editing Word problem for groups (section)

==Algebraic structure and the word problem==
There are a number of results that relate solvability of the word problem and algebraic structure. The most significant of these is the [[Boone-Higman theorem]]:

::A finitely presented group has solvable word problem if and only if it can be embedded in a [[simple group]] that can be embedded in a finitely presented group.

It is widely believed that it should be possible to do the construction so that the simple group itself is finitely presented. If so one would expect it to be difficult to prove as the mapping from presentations to simple groups would have to be non-recursive.

The following has been proved by [[Bernhard Neumann]] and [[Angus Macintyre]]:

::A finitely presented group has solvable word problem if and only if it can be embedded in every [[algebraically closed group]].

What is remarkable about this is that the algebraically closed groups are so wild that none of them has a recursive presentation.

The oldest result relating algebraic structure to solvability of the word problem is Kuznetsov's theorem:

::A recursively presented simple group <math>S</math> has solvable word problem.

To prove this let <math>\langle X | R \rangle</math> be a recursive presentation for <math>S</math>. Choose a nonidentity element <math>a \in S</math>, that is, <math>a \neq 1</math> in <math>S</math>.

If <math>w</math> is a word on the generators <math>X</math> of <math>S</math>, then let:

::<math>S_w = \langle X | R\cup \{w\} \rangle.</math>

There is a recursive function <math>f_{\langle X | R\cup \{w\} \rangle}</math> such that:

::<math>f_{\langle X | R\cup \{w\} \rangle}(x) = 
\begin{cases} 
0 &\text{if}\  x=1\ \text{in}\ S_w\\
\text{undefined/does not halt}\ &\text{if}\ x\neq 1\ \text{in}\ S_w.
\end{cases}</math>

Write:

::<math>g(w, x) = f_{\langle X | R\cup \{w\} \rangle}(x).</math>

Then because the construction of <math>f</math> was uniform, this is a recursive function of two variables.

It follows that: {{tmath|1=h(w)=g(w, a)}} is recursive. By construction:

::<math>h(w) = 
\begin{cases}
0 &\text{if}\  a=1\ \text{in}\ S_w\\
\text{undefined/does not halt}\ &\text{if}\ a\neq 1\ \text{in}\ S_w.
\end{cases}</math>

Since <math>S</math> is a simple group, its only quotient groups are itself and the trivial group. Since <math>a \neq 1</math> in <math>S</math>, we see <math>a = 1</math> in <math>S_w</math> if and only if <math>S_w</math> is trivial if and only if <math>w \neq 1</math> in <math>S</math>. Therefore:

::<math>h(w) = 
\begin{cases}
0 &\text{if}\  w\ne 1\ \text{in}\ S\\
\text{undefined/does not halt}\ &\text{if}\ w=1\ \text{in}\ S.
\end{cases}</math>

The existence of such a function is sufficient to prove the word problem is solvable for <math>S</math>.

This proof does not prove the existence of a uniform algorithm for solving the word problem for this class of groups. The non-uniformity resides in choosing a non-trivial element of the simple group. There is no reason to suppose that there is a recursive function that maps a presentation of a simple groups to a non-trivial element of the group. However, in the case of a finitely presented group we know that not all the generators can be trivial (Any individual generator could be, of course). Using this fact it is possible to modify the proof to show:

:The word problem is uniformly solvable for the class of finitely presented simple groups.