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.

Definitions used

Permutable subgroup

Further information: Permutable subgroup

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

Related facts

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. 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. Baer norm is hereditarily permutable

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

$A:=⟨a,b\mid a^{p^2}=b^p=e,ab=b{a}^{p+1}⟩$

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

We claim that the subgroup $B=⟨b⟩$ 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 ${\mathrm{\Omega }}_1(A)$ (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, $a^p\in {\mathrm{\Omega }}_1(A),b\in {\mathrm{\Omega }}_1(A)$, so ${\mathrm{\Omega }}_1(A)$ 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 ${\mathrm{\Omega }}_1(A)$, 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 ${\mathrm{\Omega }}_1(A)$, 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 ${\mathrm{\Omega }}_1(A)$ (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, $a^p\in {\mathrm{\Omega }}_1(A),b\in {\mathrm{\Omega }}_1(A)$, so ${\mathrm{\Omega }}_1(A)$ 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 ${\mathrm{\Omega }}_1(A)$ (again by fact (1)). Thus, every normalizer is either the whole group or ${\mathrm{\Omega }}_1(A)$. Moreover, since $B$ is not normal in $G$, the normalizer of $B$ is ${\mathrm{\Omega }}_1(A)$, so ${\mathrm{\Omega }}_1(A)$ is the Baer norm. Thus, since $B\le {\mathrm{\Omega }}_1(A)$, 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);
<pc group of size 9 with 2 generators>
gap> Q := SylowSubgroup(AutomorphismGroup(P),3);
<group>
gap> A := SemidirectProduct(Q,P);
<pc group with 3 generators>
gap> B := Image(Embedding(A,1));
Group([ f1 ])
gap> IsSubgroup(A,B);
true
gap> IsNormal(A,B);
false
gap> IsPermutable(A,B);
true