Characteristic not implies fully invariant in finitely generated abelian group

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a finitely generated Abelian group. That is, it states that in a finitely generated Abelian group, every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., fully characteristic subgroup)
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Statement

We can have a finitely generated abelian group with a subgroup that is characteristic but not fully invariant.

Definitions used

Finitely generated Abelian group

Further information: Finitely generated Abelian group

A finitely generated Abelian group is an Abelian group that has a finite generating set. Equivalently, it is a group expressible as a direct sum of finitely many cyclic groups.

Characteristic subgroup

Further information: Characteristic subgroup

A subgroup H of a group G is termed characteristic in G if, for every automorphism \sigma of G, \sigma(H) = H.

Fully characteristic subgroup

Further information: Fully invariant subgroup

A subgroup H of a group G is termed fully characteristic in G if, for every endomorphism \alpha of G, \alpha(H) \le H.

Related facts

Proof

There is essentially only one kind of counterexample. We describe this counterexample for the infinite case. A finite version of this counterexample can be found at characteristic not implies fully invariant in finite abelian group.

Let G be the direct sum of the infinite cyclic group and the cyclic group of order two:

G := \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}.

Let H be the cyclic subgroup of G generated by (2,1).

The subgroup is characteristic

Set:

A := \{g \in G \mid \exists x \in G, 4x = 2g \} = \{ (2r,s) \mid r \in \mathbb{Z}, s \in \mathbb{Z}/2\mathbb{Z} \}

and

B := \{g \in G \mid \exists x \in G, g = 2x \} = \{ (2r,0) \mid r \in \mathbb{Z} \}.

and:

C := \{ g \in G \mid \exists x \in G, 8x = 2g \} = \{ (4r,s) \mid r \in \mathbb{Z}, s \in \mathbb{Z}/2\mathbb{Z} \}.

Thus, we have:

D := A \setminus (B \cup C) = \{ (2r,1) \mid r \in \mathbb{Z}, r \notin 2\mathbb{Z} \}.

Clearly, any automorphism of G sends A to itself, sends B to itself, and sends C to itself. Thus, any automorphism of G sends A \setminus (B \cup C) to itself. Thus, any automorphism of G sends D to itself. Note that D comprises precisely those elements of H that have the second coordinate equal to 1: in particular, (2,1) \in D \subseteq H, so the subgroup generated by D equals H. Thus, any automorphism of G preserves H.

The subgroup is not fully invariant

Consider the map:

\pi_1:G \to G, \qquad \pi_1(a,b) = (a,0).

This map is an endomorphism of G, but the image of (2,1) under this map is (2,0), which is not an element of H. Thus, H is not fully characteristic in G.