Characteristic not implies sub-isomorph-free in finite 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 group. That is, it states that in a finite group, every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., sub-isomorph-free subgroup)
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Statement

We can have a finite group G and a characteristic subgroup H of G such that H is not a sub-isomorph-free subgroup of G. In other words, there is no ascending chain of subgroups starting at H and ending at G with each member an isomorph-free subgroup of its successor.

Related facts

Proof

The center of a non-abelian group of odd prime cube order

Further information: Prime-cube order group:U3p, Prime-cube order group:p2byp

Let p be an odd prime. Let G be the non-abelian group of order p^3 and exponent p. Let H be the center of G. Then, we have:

  • H is characteristic in G.
  • H is not sub-isomorph-free in G: In fact, no proper subgroup of G containing H is isomorph-free. The subgroups of order p^2 are all elementary abelian, while H itself is isomorphic to many other cyclic groups of prime order.

Another example is to take G as the non-abelian group of order p^3 and exponent p^2, and H to be the center of G. In this case:

  • H is characteristic in <mah>G</math>.
  • H is not sub-isomorph-free in G: H is not isomorph-free, because there are other cyclic groups of order p. There is only one isomorph-free proper subgroup of G properly containing H, namely, an elementary abelian subgroup of order p^2. But H is not isomorph-free in this subgroup.