Permutable not implies normal

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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., permutable subgroup) need not satisfy the second subgroup property (i.e., normal subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about permutable subgroup|Get more facts about normal subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property permutable subgroup but not normal subgroup|View examples of subgroups satisfying property permutable subgroup and normal subgroup

Statement

Property-theoretic statement

The subgroup property of being a permutable subgroup is not stronger than the subgroup property of being a normal subgroup.

Verbal statement

There exist situations where a permutable subgroup of a group is not normal.

Facts used

Proof

Example of a group of prime power order

Let p be an odd prime. Let A be a group generated by a,b, with the relations a^{p^2} = b^p = e and ab = ba^{p+1}. In other words, A is a semidirect product of a cyclic group of order p^2, by a cyclic group of order p. Thus, A has order p^3.

We claim that the subgroup B = \{ b \} is a permutable subgroup of A. For this, we need to show that B permutes with every cyclic subgroup of A.

Let's prove this. By the fact stated above, the set of elements of A that have order p (along with the identity element) form a subgroup of A. This subgroup (we call it D) contains a^p and b. However, it cannot be bigger than the subgroup generated by a^p and b, which itself has order p^2.

Now, if we pick any element inside D, the cyclic subgroup generated by it clearly permutes with B, because they're both inside an Abelian group of order p^2. If we pick an element outside D, then the cyclic subgroup it generates has order p^2. Hence, that cyclic subgroup is maximal in A and hence normal, so it commutes with B. Thus, every cyclic subgroup commutes with B, and we are done.