Pronormal implies self-conjugate-permutable: Difference between revisions

Revision as of 14:29, 22 October 2008

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., pronormal subgroup) must also satisfy the second subgroup property (i.e., self-conjugate-permutable subgroup)
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Statement

Property-theoretic statement

The subgroup property of being pronormal is stronger than the subgroup property of being self-conjugate-permutable, for finite groups.

Verbal statement

Any pronormal subgroup is self-conjugate-permutable.

Facts used

Product of conjugates is proper: If $G$ is a group and $H$ is a subgroup such that there exists $g\in G$ for which $HH^{g}=G$ , then $H=G$ .

Proof

Given: A group $G$ , a pronormal subgroup $H$ .

To prove: If $HH^{g}=H^{g}H$ for some $g\in G$ , then $H^{g}=H$ .

Proof: Let $K=\langle H,H^{g}\rangle =HH^{g}$ .

(Given data used: $H$ is pronormal in $G$ ): There exists $x\in K$ such that $H^{g}=H^{x}$ . Thus, $K=HH^{x}$ .
(Fact used: fact (1), product of conjugates is proper): $K$ is the product of two conjugate subgroups, so fact (1) forces that $K=H$ . Since $x\in K$ , we also get $K=H^{x}$ . Thus, $H=H^{g}$ .

@@ Line 1: / Line 1: @@
 {{subgroup property implication|
-group property = finite group|
 stronger = pronormal subgroup|
 weaker = self-conjugate-permutable subgroup}}
@@ Line 12: / Line 11: @@
 ===Verbal statement===
-Any [[pronormal subgroup]] of a [[finite group]] is [[self-conjugate-permutable subgroup|self-conjugate-permutable]].
+Any [[pronormal subgroup]]  is [[self-conjugate-permutable subgroup|self-conjugate-permutable]].
+==Facts used==
+# [[uses::Product of conjugates is proper]]: If <math>G</math> is a group and <matH>H</math> is a subgroup such that there exists <math>g \in G</math> for which <math>HH^g = G</math>, then <math>H = G</math>.
 ==Proof==
-Suppose <math>H</math> is a [[pronormal subgroup]] of a [[finite group]] <math>G</math>. Then, suppose <math>g \in G</math> is such that <math>HH^g = H^gH</math>. Clearly <math>K = HH^g</math> is a [[subgroup]] of <math>G</math>. By pronormality of <math>H</math>, there exists <math>x \in HH^g</math> such that <math>H^g = H^x</math>. Thus, <math>K</math> is a product of two conjugate subgroups inside <math>K</math>. If <math>H</math> is a ''proper'' subgroup of <math>K</math>, this contradicts the fact that a [[product of conjugates is proper]] in a finite group. Thus, <math>H = K</math>, so <math>H = H^g</math>.
+'''Given''': A group <math>G</math>, a pronormal subgroup <math>H</math>.
+'''To prove''': If <math>HH^g = H^gH</math> for some <math>g \in G</math>, then <math>H^g = H</math>.
+'''Proof''': Let <math>K = \langle H, H^g \rangle = HH^g</math>.
+# ('''Given data used''': <math>H</math> is pronormal in <math>G</math>): There exists <math>x \in K</math> such that <math>H^g = H^x</math>. Thus, <math>K = HH^x</math>.
+# ('''Fact used''': fact (1), product of conjugates is proper): <math>K</math> is the product of two conjugate subgroups, so fact (1) forces that <math>K = H</math>. Since <math>x \in K</math>, we also get <math>K = H^x</math>. Thus, <math>H = H^g</math>.