Classification of fully characteristic subgroups in finitely generated Abelian groups
Contents
Statement
Case of groups of prime power order
Suppose is an Abelian group whose order is a power of a prime
. Then, we can write, by the structure theorem for finitely generated Abelian groups:
where:
.
where .
Then a subgroup of
is a fully characteristic subgroup of
, and this happens if and only if the following are satisfied:
-
is the direct sum of the intersections
.
- If the orders of
are
, then
and
.
Case of finitely generated groups
Suppose is a finitely generated Abelian group. Then, we can write:
where is a torsion-free Abelian group, and each
is an Abelian group of prime power order for different primes. Then, a subgroup
of
is a fully characteristic subgroup, if and only if the following are satisfied:
-
is the direct sum of the intersections
.
- Each
is characteristic (equivalently, fully characteristic) in
.
Proof
Case of prime power order
Given: is an Abelian group whose order is a power of a prime
. Then, we can write, by the structure theorem for finitely generated Abelian groups:
where:
.
where .
A subgroup of
.
To prove: is characteristic if and only if it is fully characteristic in
, if and only if
is the direct sum of
, and if the orders of the intersections are
, then
and
.
Proof: The proof relies on two important homomorphisms. For , there is an injective homomorphism:
that sends the generator on the left to times the generator on the right.
There is also a surjective homomorphism:
that sends the generator to the generator.
Recall that we have:
.
For , define
as the endomorphism of
that sends
to
via the map:
.
and is zero elsewhere.
Similarly, define as the endomorphism of
that sends
to
via the map
and is zero elsewhere.
We are now in a position to prove the main result.
Suppose is a fully characteristic subgroup of
. We first show that
is a direct sum of
. For this, suppose the element:
.
Since is fully characteristic, it is invariant under the projections to the direct factors, which are endomorphisms. Thus, each
is in
. Thus, every element of
can be expressed as a sum of elements in
, and we get that
is a direct sum of the
s.
Finally, we need to show the conditions on the s. For this, observe that for
, we have:
- The endomorphism
sends
to itself: Thus,
injects into
via
, so
.
- The endomorphism
sends
to itself: Thus,
induces a surjective endomorphism from
to
, forcing
.
Now, we show that if satisfies the conditions described above, then
is fully characteristic in
.
Suppose is an endomorphism. Since
is a direct sum of
, it suffices to show that
. For this, in turn, it suffices to show that the
coordinate of
is contained in
. This is easily done in three cases:
-
: In this case, it is direct since
is a fully characteristic subgroup of the cyclic group
(all subgroups of cyclic groups are fully characteristic).
-
: In this case,
. Consider the homomorphism from
to
obtained by composing the
projection with
. This map must send
to a subgroup of size at most
in
. But since
, this subgroup of size
is contained in the subgroup
that has order
.
-
: In this case,
. COnsider the homomorphism from
to
obtained by composing the
projection with
. The index of the image of
is at least the index of
in
, which is
. Thus, the image has size at most
because
. Thus, it is in
.