Fully invariant not implies verbal in finite 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 finite abelian group. That is, it states that in a finite abelian group, every subgroup satisfying the first subgroup property (i.e., fully invariant subgroup) need not satisfy the second subgroup property (i.e., verbal subgroup)
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Statement

A fully characteristic subgroup of a finite abelian group need not be a verbal subgroup.

Related facts

Proof

Example of a group of prime-cube order

Let C_p and C_{p^2} denote the cyclic groups of order p and p^2 respectively. Let G = C_p \times C_{p^2}. Consider the omega-1 subgroup:

\Omega_1(G) := \{ x \in G \mid px = 0 \}.

In other words, it is the subgroup of G comprising the elements of order dividing p. Then, \Omega_1(G) is fully characteristic: any endomorphism of G preserves the condition. On the other hand, \Omega_1(G) is not verbal: the only possible verbal subgroups of G are the agemo subgroups: the whole group G, the set of multiples of p, and the trivial subgroup. None of these equals \Omega_1(G).