Normal-isomorph-free not implies isomorph-free in finite

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal-isomorph-free subgroup) need not satisfy the second subgroup property (i.e., isomorph-free subgroup)
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Statement

Verbal statement

We may have a group with a normal subgroup such that there is no other normal subgroup isomorphic to it, but there are other non-normal subgroups isomorphic to it.

Related facts

Proof

An example involving the symmetric group

Further information: symmetric group:S3

Let G be the symmetric group on three letters and K be the cyclic group of order two.

  • 1 \times K is a normal subgroup of G \times K. Further, if L is cyclic of order two and normal in G \times K, then the projection of L in G is normal in G. Since G has no normal subgroups of order two, the projection of L in G is trivial, so L = 1 \times K. Thus, 1 \times K is a normal-isomorph-free subgroup of G.
  • On the other hand, 1 \times K is not isomorph-free in G \times K: We can find a two-element subgroup H of G, and we then have H \times 1 \cong 1 \times K.

An example involving the dihedral group

Further information: dihedral group:D8

Let G be the dihedral group of order eight:

G = \langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle.

Let H be the center of G, so H = \langle a^2 \rangle. Then:

  • H is normal-isomorph-free: It is the only normal subgroup of order two.
  • H is not isomorph-free: There are other subgroups of order two, such as \langle x \rangle.

A generic example

For a group of prime power order, the following is true: if the center is of prime order, it is normal-isomorph-free. This follows from the fact that nilpotent implies center is normality-large: in a nilpotent group, the intersection between the center and any nontrivial normal subgroup is nontrivial. On the other hand, the center is rarely isomorph-free (an example where it is isomorph-free is the generalized quaternion group).

More generally, if P is a p-group and \Omega_1(Z(P)) denotes the set of elements of order p, then if \Omega_1(Z(P)) is cyclic of order p, it is normal-isomorph-free.