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) need not satisfy the second subgroup property (i.e., normal subgroup)
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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 $p$-group of nilpotence class two with odd order, the set of elements of order $p$ forms a subgroup (along with the identity element).

Proof

Example of a group of prime power order

Further information: prime-cube order group:p2byp Let $p$ be an odd prime. Let $A$ be the semidirect product of a cyclic group of order $p^2$ and a cyclic group of order $p$. In other words: $A$ is a group generated by $a,b$, with the relations $a^{p^2}=b^p=e$ and $ab=b{a}^{p+1}$. 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.