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 of a group
is termed permutable if
permutes with every cyclic subgroup of
.
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
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).
- Baer norm is hereditarily permutable
Proof
Example of a group of prime power order
Further information: prime-cube order group:p2byp
Let be an odd prime. Let
be the semidirect product of a cyclic group of order
and a cyclic group of order
acting nontrivially. In other words:
is a group generated by
, with the relations
and
:
Thus, has order
.
We claim that the subgroup is a permutable subgroup of
but is not normal in
. The fact that
is not normal in
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 (the subgroup generated by elements of order
) is a subgroup of exponent
(fact (1)). It cannot equal
, since
. Also,
, so
has order at least
. Hence, its order is exactly
, and it is elementary Abelian on
and
.
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.
Second proof: First, note that (the subgroup generated by elements of order
) is a subgroup of exponent
(fact (1)). It cannot equal
, since
. Also,
, so
has order at least
. Hence, its order is exactly
, and it is elementary Abelian on
and
.
By fact (2), it suffices to show that is contained in the Baer norm of
: the intersection of normalizers of all subgroups. Note that the normalizer of any subgroup of order
or
is the whole group. The normalizer of any subgroup of order
must contain that subgroup, and also the center, which is generated by
. Hence, either the subgroup is normal, or its normalizer is a subgroup of order
containing two permuting subgroups of order
, forcing it to be
(again by fact (1)). Thus, every normalizer is either the whole group or
. Moreover, since
is not normal in
, the normalizer of
is
, so
is the Baer norm. Thus, since
, we obtain that
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