Automorph-conjugacy is transitive: Difference between revisions

Revision as of 16:26, 19 December 2014

This article gives the statement, and possibly proof, of a subgroup property (i.e., automorph-conjugate subgroup) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about automorph-conjugate subgroup |Get facts that use property satisfaction of automorph-conjugate subgroup | Get facts that use property satisfaction of automorph-conjugate subgroup|Get more facts about transitive subgroup property

Statement

Suppose $H \leq K \leq G$ are groups such that $H$ is an automorph-conjugate subgroup of $K$ , and $K$ is an automorph-conjugate subgroup of $G$ . Then, $H$ is an automorph-conjugate subgroup of $G$ .

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Given: Groups $H \leq K \leq G$ such that $H$ is an automorph-conjugate subgroup of $K$ and $K$ is an automorph-conjugate subgroup of $G$ . An automorphism $σ$ of $G$ .

To prove: There exists $x \in G$ such that $σ (H) = x H x^{- 1}$

Proof:

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	$σ (H) \leq σ (K)$	$H \leq K$		given-direct
2	There exists $g \in G$ such that $g K g^{- 1} = σ (K)$	$K$ is an automorph-conjugate subgroup of $G$ $σ$ is an automorphism of $G$		given-direct
3	Denote by $c_{g}$ the map $u \mapsto g u g^{- 1}$ . Then $c_{g}^{- 1} \circ s i g m a$ is an automorphism of $G$ that restricts to an automorphism of $K$ .		Step (2)	direct from the step
4	There exists $k \in K$ such that $(c_{g}^{- 1} \circ σ) (H) = k H k^{- 1}$ .	$H$ is an automorph-conjugate subgroup of $K$	Step (3)	Step-given combination direct
5	Setting $x = g k$ , we get that $σ (H) = x H x^{- 1}$		Step (4)	Simple algebraic manipulation gives $σ (H) = c_{g} (k H k^{- 1})$ $= g k H k^{- 1} g^{- 1} = (g k) H (g k)^{- 1}$

@@ Line 8: / Line 8: @@
 ==Proof==
+{{tabular proof format}}
 '''Given''': Groups <math>H \le K \le G</math> such that <math>H</math> is an automorph-conjugate subgroup of <math>K</math> and <math>K</math> is an automorph-conjugate subgroup of <math>G</math>. An automorphism <math>\sigma</math> of <math>G</math>.