# Permutable not implies normal

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) neednotsatisfy 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

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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

## Contents

## 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

- Omega-1 of odd-order class two p-group has prime exponent: For a -group of nilpotence class two with odd order, the set of elements of order forms a subgroup (along with the identity element).

## Proof

### Example of a group of prime power order

Let be an odd prime. Let be a group generated by , with the relations and . In other words, is a semidirect product of a cyclic group of order , by a cyclic group of order . Thus, has order .

We claim that the subgroup is a permutable subgroup of . For this, we need to show that permutes with every *cyclic* subgroup of .

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

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