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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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.
- Characteristic not implies characteristic-isomorph-free
- Characteristic-isomorph-free not implies normal-isomorph-free
An example involving the symmetric group
Further information: symmetric group:S3
Let be the symmetric group on three letters and be the cyclic group of order two.
- is a normal subgroup of . Further, if is cyclic of order two and normal in , then the projection of in is normal in . Since has no normal subgroups of order two, the projection of in is trivial, so . Thus, is a normal-isomorph-free subgroup of .
- On the other hand, is not isomorph-free in : We can find a two-element subgroup of , and we then have .
An example involving the dihedral group
Further information: dihedral group:D8
Let be the dihedral group of order eight:
Let be the center of , so . Then:
- is normal-isomorph-free: It is the only normal subgroup of order two.
- is not isomorph-free: There are other subgroups of order two, such as .
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 is a -group and denotes the set of elements of order , then if is cyclic of order , it is normal-isomorph-free.