Automorph-conjugacy is transitive: Difference between revisions

Latest revision as of 03:08, 12 January 2026

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)
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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

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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 $c_{g}^{- 1} \circ σ$ 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 1: / Line 1: @@
-{{subgroup metaproperty satisfaction}}
+{{subgroup metaproperty satisfaction|
+property = automorph-conjugate subgroup|
+metaproperty = transitive subgroup property}}
 ==Statement==
-===Property-theoretic statement===
+Suppose <math>H \le K \le G</math> are groups 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>. Then, <math>H</math> is an automorph-conjugate subgroup of <math>G</math>.
-The [[subgroup property]] of being [[automorph-conjugate subgroup|automorph-conjugate]] is [[transitive subgroup property|transitive]].
+==Proof==
-===Symbolic statement===
+{{tabular proof format}}
-Let <math>H</math> be an automorph-conjugate subgroup of <math>K</math>, and <math>K</math> be an automorph-conjugate subgroup of <math>G</math>. Then, <math>H</math> is an automorph-conjugate subgroup of <math>G</math>.
+'''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>.
-==Proof==
+'''To prove''': There exists <math>x \in G</math> such that <math>\sigma(H) = xHx^{-1}</math>
-Suppose <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>. We want to show that <math>H</math> is an automorph-conjugate subgroup of <math>G</math>.
+'''Proof''':
-For this, pick any automorphism <math>\sigma</math> of <math>H</math>. Clearly, <math>\sigma(H) \le \sigma(K)</math>, and since <math>K</math> is automorph-conjugate subgroup of <math>G</math>, there exists <math>g \in G</math> such that <math>g \sigma(K) g^{-1} = K</math>. Thus, <math>c_g \circ \sigma</math> (conjugation by <math>g</math>, composed with <math>\sigma</math>), gives an automorphism of <math>K</math>. Since <math>H</math> is automorph-conjugate inside <math>K</math>, there exists <math>h \in K</math> such that <math>g \sigma(H) g^{-1} = hHh^{-1}</math>. Rearranging, we see that <math>\sigma(H) = g^{-1}h H h^{-1}g</math>, a conjugate fo <math>H</math>.
+{| class="sortable" border="1"
+! Step no. || Assertion/construction !! Given data used !! Previous steps used !! Explanation
+|-
+| 1 || <math>\sigma(H) \le \sigma(K)</math> || <math>H \le K</math> || || given-direct
+|-
+| 2 || There exists <math>g \in G</math> such that <math>g K g^{-1} = \sigma(K)</math> ||<math>K</math> is an automorph-conjugate subgroup of <math>G</math><br><math>\sigma</math> is an automorphism of <math>G</math> || || given-direct
+|-
+| 3 || Denote by <math>c_g</math> the map <math>u \mapsto gug^{-1}</math>. Then <math>c_g^{-1} \circ \sigma</math> is an automorphism of <math>G</math> that restricts to an automorphism of <math>K</math>. || || Step (2) || direct from the step
+|-
+| 4 || There exists <math>k \in K</math> such that <math>(c_g^{-1} \circ \sigma)(H) = kHk^{-1}</math>. || <math>H</math> is an automorph-conjugate subgroup of <math>K</math> || Step (3) || Step-given combination direct
+|-
+| 5 || Setting <math>x = gk</math>, we get that <math>\sigma(H) = xHx^{-1}</math>  || || Step (4) || Simple algebraic manipulation gives <math>\sigma(H) = c_g(kHk^{-1})</math><br><math>= gkHk^{-1}g^{-1} = (gk)H(gk)^{-1}</math>
+|}