In mathematics, in the realm of group theory, a group
is algebraically closed if any finite set of equations and inequations that "make sense" in
already have a solution in
. This idea will be made precise later in the article.
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Suppose we wished to find an element
of a group
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.)
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Now suppose
is the group with the multiplication table:
Then the conditions:


have a solution in
, namely
.
However the conditions:


Do not have a solution in
, as can easily be checked.
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However if we extend the group
to the group
with multiplication table:
Then the conditions have two solutions, namely
and
.
Thus there are three possibilities regarding such conditions:
and have no solution in any extension of
.
.
but nevertheless have a solution in some extension
of
.It is reasonable to ask whether there are any groups
such that whenever a set of conditions like these have a solution at all, they have a solution in
itself? The answer turns out to be "yes", and we call such groups algebraically closed groups.
We first need some preliminary ideas.
If
is a group and
is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in
we mean a pair of subsets
and
of
the free product of
and
.
This formalizes the notion of a set of equations and inequations consisting of variables
and elements
of
. The set
represents equations like:



The set
represents inequations like


By a solution in
to this finite set of equations and inequations, we mean a homomorphism
, such that
for all
and
for all
. Where
is the unique homomorphism
that equals
on
and is the identity on
.
This formalizes the idea of substituting elements of
for the variables to get true identities and inidentities. In the example the substitutions
and
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
if there is a group
and an embedding
such that the finite set of equations and inequations
and
has a solution in
. Where
is the unique homomorphism
that equals
on
and is the identity on
.
Now we formally define the group
to be algebraically closed if every finite set of equations and inequations that has coefficients in
and is consistent with
has a solution 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.
First we embed
in a countable group
with the property that every finite set of equations with coefficients in
that is consistent in
has a solution in
as follows:
There are only countably many finite sets of equations and inequations with coefficients in
. Fix an enumeration
of them. Define groups
inductively by:


Now let:

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

Then
is a countable group containing
. It is algebraically closed because any finite set of equations and inequations that is consistent with
must have coefficients in some
and so must have a solution in
.
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