Normal not implies potentially fully invariant: Difference between revisions

Latest revision as of 19:45, 11 August 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., normal subgroup) need not satisfy the second subgroup property (i.e., potentially fully invariant subgroup)
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Statement

It is possible to have a normal subgroup $H$ of a group $G$ that is not a potentially fully invariant subgroup of $G$ -- in other words, there is no group $K$ containing $G$ such that $H$ is a fully invariant subgroup of $K$ .

Related facts

Facts used

Equivalence of definitions of complete direct factor: This states that for a complete subgroup, being normal is equivalent to being a direct factor.
Equivalence of definitions of fully invariant direct factor: This states that for a direct factor, being a fully invariant subgroup is equivalent to being a homomorph-containing subgroup.
Homomorph-containment satisfies intermediate subgroup condition

Proof

Example involving a complete group

Further information: Complete and potentially fully invariant implies homomorph-containing

Let $A$ be a nontrivial complete group. Define $G : = A \times A$ and $H : = A \times {e}$ . Clearly, $H$ is a normal subgroup of $G$ .

Suppose $K$ is a group containing $G$ , such that $H$ is fully invariant in $K$ . In particular, $H$ is normal in $K$ . Since $H$ is complete, it is a direct factor, so there exists a group $C$ that is a complement to $H$ , so $K = H \times C$ as an internal direct product. Further, since $G / H ≅ H ≅ A$ is a subgroup of $K / H$ , $C$ has a subgroup, say $B$ , isomorphic to $A ≅ H$ .

Then, consider the endomorphism $α$ of $K$ that sends $C$ to the trivial subgroup and $H$ isomorphically to the subgroup $B$ . This endomorphism does not send $H$ to within itself.

More general example

Further information: Fully normalized and potentially fully invariant implies centralizer-annihilating endomorphism-invariant

More generally, suppose $H$ is a fully normalized subgroup of $G$ that is normal in $G$ , but such that there is a homomorphism $θ : G \to G$ whose kernel contains $C_{G} (H)$ such that $θ (H)$ is not contained in $H$ (in other words, $H$ is not a centralizer-annihilating endomorphism-invariant subgroup). Then, $H$ is not a potentially fully invariant subgroup of $G$ .

Examples include:

$G$ is the dihedral group of order 16, say $G : = ⟨ a, x ∣ a^{8} = x^{2} = e, x a x = x^{- 1} ⟩$ , and $H = ⟨ a^{2}, x ⟩$ . Then, $C_{G} (H) = ⟨ a^{4} ⟩$ and we have a homomorphism $θ$ from $G$ to $G$ such that $θ (a) = a^{2}, θ (x) = a x$ with kernel containing $C_{G} (H)$ and with $θ (H)$ not contained in $H$ . Thus, $H$ is not a potentially fully invariant subgroup of $G$ . Further information: D8 is not potentially fully invariant in D16

@@ Line 9: / Line 9: @@
 ==Related facts==
+* [[Normal not implies image-potentially fully invariant]]
 * [[Normal not implies potentially verbal]]
 * [[NPC theorem]]: Normal equals potentially ''characteristic''.
 * [[Normal equals potentially normal-subhomomorph-containing]]
+* [[Complete and potentially fully invariant implies homomorph-containing]]
+* [[Fully normalized and potentially fully invariant implies centralizer-annihilating endomorphism-invariant]]
+==Facts used==
+# [[uses::Equivalence of definitions of complete direct factor]]: This states that for a complete subgroup, being normal is equivalent to being a direct factor.
+# [[uses::Equivalence of definitions of fully invariant direct factor]]: This states that for a [[direct factor]], being a [[fully invariant subgroup]] is equivalent to being a [[homomorph-containing subgroup]].
+# [[uses::Homomorph-containment satisfies intermediate subgroup condition]]
 ==Proof==
 ===Example involving a complete group===
+{{further|[[Complete and potentially fully invariant implies homomorph-containing]]}}
 Let <math>A</math> be a nontrivial [[complete group]]. Define <math>G := A \times A</math> and <math>H := A \times \{ e \}</math>. Clearly, <math>H</math> is a normal subgroup of <math>G</math>.
@@ Line 21: / Line 32: @@
 Suppose <math>K</math> is a group containing <math>G</math>, such that <math>H</math> is fully invariant in <math>K</math>. In particular, <math>H</math> is normal in <math>K</math>. Since <math>H</math> is complete, it is a direct factor, so there exists a group <math>C</math> that is a complement to <math>H</math>, so <math>K = H \times C</math> as an internal direct product. Further, since <math>G/H \cong H \cong A</math> is a subgroup of <math>K/H</math>, <math>C</math> has a subgroup, say <math>B</math>, isomorphic to <math>A \cong H</math>.
-Then, consider the endomorphism <math>\alpha</math> of <math>K</math> that sends <math>C</math> to the trivial subgroup and <math>H</math> isomorphically to the subgroup <math>B</math> does ''not'' send <math>H</math> to within itself.
+Then, consider the endomorphism <math>\alpha</math> of <math>K</math> that sends <math>C</math> to the trivial subgroup and <math>H</math> isomorphically to the subgroup <math>B</math>. This endomorphism does ''not'' send <math>H</math> to within itself.
 ===More general example===
@@ Line 27: / Line 38: @@
 {{further|[[Fully normalized and potentially fully invariant implies centralizer-annihilating endomorphism-invariant]]}}
-More generally, suppose <math>H</math> is a [[fully normalized subgroup]] of <math>G</math> that is [[normal subgroup|normal]] in <math>G</math>, but such that there is a homomorphism <math>\theta: G/C_G(H) \to G</math> such that <math>\theta(H)</math> is not contained in <math>H</math> (in other words, <math>H</math> is not a [[centralizer-annihilating endomorphism-invariant subgroup]]).
+More generally, suppose <math>H</math> is a [[fully normalized subgroup]] of <math>G</math> that is [[normal subgroup|normal]] in <math>G</math>, but such that there is a homomorphism <math>\theta: G \to G</math> whose kernel contains <math>C_G(H)</math> such that <math>\theta(H)</math> is not contained in <math>H</math> (in other words, <math>H</math> is not a [[centralizer-annihilating endomorphism-invariant subgroup]]).
 Then, <math>H</math> is ''not'' a [[potentially fully invariant subgroup]] of <math>G</math>.
+Examples include:
+* <math>G</math> is the [[dihedral group:D16|dihedral group of order 16]], say <math>G := \langle a,x \mid a^8 = x^2 = e, xax = x^{-1}\rangle</math>, and <math>H = \langle a^2,x \rangle</math>. Then, <math>C_G(H) = \langle a^4 \rangle</math> and we have a homomorphism <math>\theta</math> from <math>G</math> to <math>G</math> such that <math>\theta(a) = a^2, \theta(x) = ax</math> with kernel containing <math>C_G(H)</math> and with <math>\theta(H)</math> not contained in <math>H</math>. Thus, <math>H</math> is not a potentially fully invariant subgroup of <math>G</math>. {{further|[[D8 is not potentially fully invariant in D16]]}}