Permutable not implies normal

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.

Permutable subgroup
A subgroup $$H$$ of a group $$G$$ is termed permutable if $$H$$ permutes with every cyclic subgroup of $$G$$.

Similarities in behavior between permutability and normality
Get more information at permutable versus normal.


 * Permutability is strongly join-closed
 * Permutability satisfies intermediate subgroup condition
 * Permutability satisfies transfer condition
 * Permutability satisfies intermediate subgroup condition
 * Permutability satisfies image condition

Differences between permutability and normality

 * Permutability is not intersection-closed

Other related facts

 * Baer norm is hereditarily permutable
 * Baer norm not is hereditarily normal

Facts used

 * 1) uses::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).
 * 2) uses::Baer norm is hereditarily permutable

Example of a group of prime power order
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 $$p$$ acting nontrivially. In other words: $$A$$ is a group generated by $$a,b$$, with the relations $$a^{p^2} = b^p = e$$ and $$ab = ba^{p+1}$$:

$$A := \langle a,b \mid a^{p^2} = b^p = e, ab = ba^{p+1} \rangle$$

Thus, $$A$$ has order $$p^3$$.

We claim that the subgroup $$B = \langle b \rangle$$ is a permutable subgroup of $$A$$ but is not normal in $$A$$. The fact that $$B$$ is not normal in $$A$$ is direct from the fact that it is the acting group in a semidirect product corresponding to a nontrivial action. For permutability, we present two (essentially equivalent) forms of the proof.

First proof: First, note that $$\Omega_1(A)$$ (the subgroup generated by elements of order $$p$$) is a subgroup of exponent $$p$$ (fact (1)). It cannot equal $$A$$, since $$a^p \ne e$$. Also, $$a^p \in \Omega_1(A), b \in \Omega_1(A)$$, so $$\Omega_1(A)$$ has order at least $$p^2$$. Hence, its order is exactly $$p^2$$, and it is elementary Abelian on $$a^p$$ and $$b$$.

Now, if we pick any element inside $$\Omega_1(A)$$, 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 $$\Omega_1(A)$$, 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.

Second proof: First, note that $$\Omega_1(A)$$ (the subgroup generated by elements of order $$p$$) is a subgroup of exponent $$p$$ (fact (1)). It cannot equal $$A$$, since $$a^p \ne e$$. Also, $$a^p \in \Omega_1(A), b \in \Omega_1(A)$$, so $$\Omega_1(A)$$ has order at least $$p^2$$. Hence, its order is exactly $$p^2$$, and it is elementary Abelian on $$a^p$$ and $$b$$.

By fact (2), it suffices to show that $$B$$ is contained in the Baer norm of $$A$$: the intersection of normalizers of all subgroups. Note that the normalizer of any subgroup of order $$1, p^2$$ or $$p^3$$ is the whole group. The normalizer of any subgroup of order $$p$$ must contain that subgroup, and also the center, which is generated by $$a^p$$. Hence, either the subgroup is normal, or its normalizer is a subgroup of order $$p^2$$ containing two permuting subgroups of order $$p$$, forcing it to be $$\Omega_1(A)$$ (again by fact (1)). Thus, every normalizer is either the whole group or $$\Omega_1(A)$$. Moreover, since $$B$$ is not normal in $$G$$, the normalizer of $$B$$ is $$\Omega_1(A)$$, so $$\Omega_1(A)$$ is the Baer norm. Thus, since $$B \le \Omega_1(A)$$, we obtain that $$B$$ is permutable.

GAP implementation
To do this implementation successfully, first define the GAP function for permutable subgroup, as available at GAP:IsPermutable. This function can be defined inline, in the interactive interface (enter it line by line) or put in a file read by the Read command.

gap> P := CyclicGroup(9);  gap> Q := SylowSubgroup(AutomorphismGroup(P),3); gap> A := SemidirectProduct(Q,P);  gap> B := Image(Embedding(A,1)); Group([ f1 ]) gap> IsSubgroup(A,B); true gap> IsNormal(A,B); false gap> IsPermutable(A,B); true