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

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

 * Characteristic not implies characteristic-isomorph-free
 * Characteristic-isomorph-free not implies normal-isomorph-free

An example involving the symmetric group
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
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.