Endo-invariance implies strongly join-closed: Difference between revisions

Latest revision as of 01:01, 16 January 2010

This article gives the statement and possibly, proof, of an implication relation between two subgroup metaproperties. That is, it states that every subgroup satisfying the first subgroup metaproperty (i.e., endo-invariance property) must also satisfy the second subgroup metaproperty (i.e., strongly join-closed subgroup property)
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Statement

Verbal statement

Any subgroup property that arises as an invariance property with respect to endomorphisms in the function restriction formalism is strongly join-closed, viz it is both join-closed and trivially true.

Symbolic statement

Let $p$ be an endomorphism property. Let $I$ be a (possibly empty) indexing set. Let $H_{i}$ is a family of subgroups of $G$ indexed by $I$ . Assume that for every function $f$ on $G$ satisfying $p$ , $f (H_{i})$ ⊆ $H_{i}$ (viz $H_{i}$ satisfies the invariance property for $p$ ).

Then, if $H$ denotes the join of (viz, subgroup generated by) all $H_{i}$ s, $H$ also satisfies the invariance property for $p$ . In other words, whenever $f$ is a function on $G$ satisfying $p$ , $f (H)$ ⊆ $H$ .

Examples

A join of normal subgroups is normal. Here, the subgroup property of being normal is the property of being invariant under inner automorphisms, and is hence the invariance property for the property of being an inner automorphism.
A join of characteristic subgroups is characteristic. Here the subgroup property of being characteristic is the invariance property with respect to the property of being an automorphism.

Definitions used

Invariance property

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Strongly join-closed subgroup property

A subgroup property is termed strongly join-closed if given any family of subgroups having the property, their join (viz the subgroup generated by them) also has the property. Note that just saying that a subgroup property is join-closed simply means that given any nonempty family of subgroups with the property, the join also has the property.

Thus, the property of being strongly intersection-closed is the conjunction of the properties of being intersection-closed and trivially true, viz satisfied by the trivial subgroup.

Facts used

Homomorphisms commute with joins: This states that if $σ : G \to K$ is a homomorphism, and $H_{i}, i \in I$ is a collection of subgroups of $G$ whose join is $H$ , then $σ (H)$ is the join of $σ (H_{i}), i \in I$ .

Proof

Given: A property $p$ of endomorphisms, a group $G$ with subgroups $H_{i}, i \in I$ whose join is a subgroup $H$ of $G$ . Further, each $H_{i}$ is invariant under each endomorphism $σ$ of $G$ that satisfies $p$ .

To prove: $H$ is invariant under each endomorphism $σ$ of $G$ that satisfies property $p$ .

Proof: We pick any endomorphism $σ$ of $G$ that satisfies $p$ .

For each $i \in I$ , $σ (H_{i})$ is contained in $H_{i}$ : This follows from the assumption on the $H_{i}$ s.
$σ (H)$ is the join of the $σ (H_{i})$ s: This follows from fact (1), with $K = G$ .
$σ (H)$ is contained in $H$ : Since each $σ (H_{i})$ is contained in $H_{i}$ , it is in particular contained in $H$ , hence their join, which is $σ (H)$ , is also contained in $H$ .

This completes the proof.

@@ Line 1: / Line 1: @@
-{{subgroup metaproperty implication}}
+{{subgroup metaproperty implication|
+stronger = endo-invariance property|
+weaker = strongly join-closed subgroup property}}
 ==Statement==
@@ Line 30: / Line 32: @@
 Thus, the property of being strongly intersection-closed is the conjunction of the properties of being intersection-closed and [[trivially true subgroup property|trivially true]], viz satisfied by the trivial subgroup.
+==Facts used==
+# [[uses::Homomorphisms commute with joins]]: This states that if <math>\sigma:G \to K</math> is a homomorphism, and <math>H_i, i \in I</math> is a collection of subgroups of <math>G</math> whose join is <math>H</math>, then <math>\sigma(H)</math> is the join of <math>\sigma(H_i), i \in I</math>.
 ==Proof==
-Let <math>p</math> be a property of endomorphisms.
+'''Given''': A property <math>p</math> of endomorphisms, a group <math>G</math> with subgroups <math>H_i, i \in I</math> whose join is a subgroup <math>H</math> of <math>G</math>. Further, each <math>H_i</math> is invariant under each endomorphism <math>\sigma</math> of <math>G</math> that satisfies <math>p</math>.
-Let <math>H_i</math> be a family of subgroups of <math>G</math> indexed by <math>I</math>, such that each <math>H_i</math> is invariant under any endomorphism of <math>G</math> satisfying <math>p</math>. Suppose <math>H</math> is the subgroup generated by them. We need to show that <math>H</math> satisfies the invariance property for <math>p</math> in <math>G</math>. In other words, for any <math>x</math> in <math>H</math> and any endomorphism <math>\sigma</math> of <math>G</math> satisfying <math>p</math>, we need to show that <math>\sigma(x)</math> is also in <math>H</math>.
+'''To prove''': <math>H</math> is invariant under each endomorphism <math>\sigma</math> of <math>G</math> that satisfies property <math>p</math>.
-The idea behind the proof is argue that since <math>\sigma</math> is an endomorphism, <math>\sigma(H)</math> is generated by the <math>\sigma(H_i)</math>. To prove this consider an arbitrary element in <math>\sigma(H)</math>. This can be written as <math>\sigma(h)</math> for some <math>h</math> in <math>H</math>. By the definition of <math>H</math>, <math>h</math> can be expressed as a product of finite length involving the elements of <math>H_i</math>s.
+'''Proof''': We pick any endomorphism <math>\sigma</math> of <math>G</math> that satisfies <math>p</math>.
-Now, since <math>\sigma</math> is an endomorphism, we can express <math>\sigma(h)</math> as the same word in the images of the corresponding elements of the <math>H_i</math>s. Hence, every element of <math>\sigma(H)</math> is generated by the elements of the <math>\sigma(H_i)</math>s.
+# For each <math>i \in I</math>, <math>\sigma(H_i)</math> is contained in <math>H_i</math>: This follows from the assumption on the <math>H_i</math>s.
+# <math>\sigma(H)</math> is the join of the <math>\sigma(H_i)</math>s: This follows from fact (1), with <math>K = G</math>.
+# <math>\sigma(H)</math> is contained in <math>H</math>: Since each <math>\sigma(H_i)</math> is contained in <math>H_i</math>, it is in particular contained in <math>H</math>, hence their join, which is <math>\sigma(H)</math>, is also contained in <math>H</math>.
-From this, it readily follows that if <math>\sigma(H_i)</math> is contained in <math>H_i</math> for each <math>i</math>, then <math>\sigma(H)</math> is contained in <math>H</math>.
+This completes the proof.