is algebraically closed if any finite set of equations and inequations that are applicable to

without needing a group extension.

This notion will be made precise later in the article in § Formal definition.

Suppose we wished to find an element

satisfying the conditions (equations and inequations): Then it is easy to see that this is impossible because the first two equations imply

In this case we say the set of conditions are inconsistent with

(In fact this set of conditions are inconsistent with any group whatsoever.)

is the group with the multiplication table to the right.

with the adjacent multiplication table: Then the conditions have two solutions, namely

Thus there are three possibilities regarding such conditions: It is reasonable to ask whether there are any groups

We first need some preliminary ideas.

is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in

This formalizes the notion of a set of equations and inequations consisting of variables

represents inequations like By a solution in

to this finite set of equations and inequations, we mean a homomorphism

This formalizes the idea of substituting elements of

for the variables to get true identities and inidentities.

yield: We say the finite set of equations and inequations is consistent with

if we can solve them in a "bigger" group

More formally: The equations and inequations are consistent with

such that the finite set of equations and inequations

Now we formally define the group

to be algebraically closed if every finite set of equations and inequations that has coefficients in

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 of the proof that a countable group

can be embedded in an algebraically closed group follows.

with the property that every finite set of equations with coefficients in

as follows: There are only countably many finite sets of equations and inequations with coefficients in

inductively by: Now let: Now iterate this construction to get a sequence of groups

It is algebraically closed because any finite set of equations and inequations that is consistent with