Characteristic not implies fully invariant in finite abelian group

From Groupprops
Revision as of 02:55, 4 July 2012 by Vipul (talk | contribs) (The subgroup is characteristic)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a finite abelian group. That is, it states that in a finite abelian group, every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., fully invariant subgroup)
View all subgroup property non-implications | View all subgroup property implications


In a finite abelian group, a characteristic subgroup need not be a fully invariant subgroup.

Related facts


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

G := \mathbb{Z}/8\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}.

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

The subgroup is characteristic


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


B := \{g \in G \mid \exists x \in G, g = 2x \} = \{ (2r,0) \mid r \in \mathbb{Z}/8\mathbb{Z} \}.


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