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

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 $H \times 1 ≅ 1 \times K$ .

An example involving the dihedral group

Further information: dihedral group:D8

Let $G$ be the dihedral group of order eight:

$G = ⟨ a, x ∣ a^{4} = x^{2} = e, x a x^{- 1} = a^{- 1} ⟩$ .

Let $H$ be the center of $G$ , so $H = ⟨ a^{2} ⟩$ . 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 $⟨ x ⟩$ .

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 $Ω_{1} (Z (P))$ denotes the set of elements of order $p$ , then if $Ω_{1} (Z (P))$ is cyclic of order $p$ , it is normal-isomorph-free.