Characteristicity is not finite direct power-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) not satisfying a subgroup metaproperty (i.e., finite direct power-closed subgroup property).
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Statement with symbols

It is possible to have a group G and a characteristic subgroup H of G such that in the direct square G \times G, the corresponding direct square H \times H, viewed as a subgroup, is not a characteristic subgroup.

Related facts

Notion of finite direct power-closed characteristic

A subgroup H of a group G is termed a finite direct power-closed characteristic subgroup if for any natural number n, H^n is characteristic in G^n.


Further information: direct product of Z8 and Z2

Suppose G = \mathbb{Z}_8 \times \mathbb{Z}_2, and H is the subgroup of order four given by:

H := \{ (2,1), (4,0), (6,1), (0,0) \}

Note that this subgroup is characteristic but not fully invariant, and is a standard example of the fact that characteristic not implies fully invariant in finite abelian group.

Then, we claim that H \times H is not a characteristic subgroup inside G \times G. To see this, consider the endomorphism of G that projects onto the \mathbb{Z}_8 coordinate, i.e., the map:

\pi: (a,b) \mapsto (a, 0)

Now, consider the automorphism of G \times G defined as:

(x,y) \mapsto (x, y + \pi(x))

This is an endomorphism as it is a sum of endomorphisms; it is an automorphism because it has an inverse:

(x,y) \mapsto (x, y - \pi(x))

In full, the automorphism is:

((a,b),(c,d)) \mapsto ((a,b),(c + a,d))

Under this automorphism, the element:

((2,1),(2,1)) \in H \times H

gets mapped to the element:


which is not inside H \times H.