Transitively normal not implies central factor: Difference between revisions

Latest revision as of 21:21, 16 January 2010

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., transitively normal subgroup) need not satisfy the second subgroup property (i.e., central factor)
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EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property transitively normal subgroup but not central factor|View examples of subgroups satisfying property transitively normal subgroup and central factor

Statement

It is possible to have a group $G$ and a transitively normal subgroup $K$ of $G$ (i.e., any normal subgroup $H$ of $K$ is normal in $G$ ) that is not a central factor of $G$ .

Proof

Example of the dihedral group

Further information: dihedral group:D8, cyclic maximal subgroup of dihedral group:D8

Consider the dihedral group of order eight:

$G : = ⟨ a, x ∣ a^{4} = x^{2} = e, x a x = a^{- 1} ⟩$ .

Suppose $K$ is the cyclic subgroup of order four:

$K : = ⟨ a ⟩$ .

Then:

$K$ is a transitively normal subgroup of $G$ : $K$ is normal (it has index two in $G$ ), and its proper subgroups (the trivial subgroup, and the subgroup $⟨ a^{2} ⟩$ ) are all normal in $G$ as well.
$K$ is not a central factor of $G$ : Conjugation by $x$ gives an automorphism of $K$ that sends $a$ to $a^{- 1}$ , and is hence not an inner automorphism of $K$ .

@@ Line 5: / Line 5: @@
 ==Statement==
-It is possible to have a [[group]] <math>G</math> and a [[transitively normal subgroup]] <math>H</math> of <math>G</math> (i.e., any [[normal subgroup]] <math>K</math> of <math>H</math> is normal in <math>G</math>) that is not a [[central factor]] of <math>G</math>.
+It is possible to have a [[group]] <math>G</math> and a [[transitively normal subgroup]] <math>K</math> of <math>G</math> (i.e., any [[normal subgroup]] <math>H</math> of <math>K</math> is normal in <math>G</math>) that is not a [[central factor]] of <math>G</math>.
 ==Proof==
@@ Line 17: / Line 17: @@
 <math>G := \langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle</math>.
-Suppose <math>H</math> is the cyclic subgroup of order four:
+Suppose <math>K</math> is the cyclic subgroup of order four:
-<math>H := \langle a \rangle</math>.
+<math>K := \langle a \rangle</math>.
 Then:
-* <math>H</math> is a transitively normal subgroup of <math>G</math>: <math>H</math> is normal (it has index two in <math>G</math>), and its proper subgroups (the trivial subgroup, and the subgroup <math>\langle a^2 \rangle</math>) are all normal in <math>G</math> as well.
+* <math>K</math> is a transitively normal subgroup of <math>G</math>: <math>K</math> is normal (it has index two in <math>G</math>), and its proper subgroups (the trivial subgroup, and the subgroup <math>\langle a^2 \rangle</math>) are all normal in <math>G</math> as well.
-* <math>H</math> is not a central factor of <math>G</math>: Conjugation by <math>x</math> gives an automorphism of <math>H</math> that sends <math>a</math> to <math>a^{-1}</math>, and is hence not an inner automorphism of <math>H</math>.
+* <math>K</math> is not a central factor of <math>G</math>: Conjugation by <math>x</math> gives an automorphism of <math>K</math> that sends <math>a</math> to <math>a^{-1}</math>, and is hence not an inner automorphism of <math>K</math>.