Automorph-conjugacy is transitive - Revision history

Vipul: /* Proof */

2026-01-12T03:08:46Z

Proof

← Older revision		Revision as of 03:08, 12 January 2026
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	\| 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		\| 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		\| 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		\| 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

Vipul: /* Proof */

2014-12-19T16:26:33Z

Proof

← Older revision		Revision as of 16:26, 19 December 2014
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	==Proof==		==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>.		'''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>.

Vipul: /* Proof */

2014-12-19T16:26:22Z

Proof

← Older revision		Revision as of 16:26, 19 December 2014
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	\| 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		\| 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 automorph-conjugate in <math>K</math> \|\| Step (3) \|\| Step-given combination direct		\| 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>		\| 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>
	\|}		\|}

Vipul: /* Proof */

2014-12-19T16:26:01Z

Proof

← Older revision		Revision as of 16:26, 19 December 2014
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	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Step no. \|\| Assertion/construction ~~!! Facts used~~ !! Given data used !! Previous steps used !! Explanation		! 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		\| 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		\| 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		\| 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 automorph-conjugate in <math>K</math> \|\| Step (3) \|\| Step-given combination direct		\| 4 \|\| There exists <math>k \in K</math> such that <math>(c_g^{-1} \circ \sigma)(H) = kHk^{-1}</math>. \|\| <math>H</math> is automorph-conjugate in <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>		\| 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>
	\|}		\|}

Vipul: /* Proof */

2014-12-19T16:25:31Z

Proof

← Older revision		Revision as of 16:25, 19 December 2014
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	\| 4 \|\| There exists <math>k \in K</math> such that <math>(c_g^{-1} \circ \sigma)(H) = kHk^{-1}</math>. \|\| \|\| <math>H</math> is automorph-conjugate in <math>K</math> \|\| Step (3) \|\| Step-given combination direct		\| 4 \|\| There exists <math>k \in K</math> such that <math>(c_g^{-1} \circ \sigma)(H) = kHk^{-1}</math>. \|\| \|\| <math>H</math> is automorph-conjugate in <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}) = gkHk^{-1}g^{-1} = (gk)H(gk)^{-1}</math>.		\| 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>
	\|}		\|}

Vipul: /* Proof */

2014-12-19T16:25:01Z

Proof

← Older revision		Revision as of 16:25, 19 December 2014
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	\| 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		\| 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 \~~maps~~ 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		\| 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 automorph-conjugate in <math>K</math> \|\| Step (3) \|\| Step-given combination direct		\| 4 \|\| There exists <math>k \in K</math> such that <math>(c_g^{-1} \circ \sigma)(H) = kHk^{-1}</math>. \|\| \|\| <math>H</math> is automorph-conjugate in <math>K</math> \|\| Step (3) \|\| Step-given combination direct

Vipul: |

2014-12-19T16:24:48Z

← Older revision		Revision as of 16:24, 19 December 2014
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	==Proof==		==Proof==

	'''Given''': <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>.		'''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>.

	'''To prove''': There exists <math>x \in G</math> such that <math>\sigma(H) = xHx^{-1}</math>		'''To prove''': There exists <math>x \in G</math> such that <math>\sigma(H) = xHx^{-1}</math>

	'''Proof''': ~~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>~~, so setting~~ <math>x = g^{-1}h</math> ~~works~~.		'''Proof''':

			{\| class="sortable" border="1"
			! Step no. \|\| Assertion/construction !! Facts used !! 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 \maps 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 automorph-conjugate in <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}) = gkHk^{-1}g^{-1} = (gk)H(gk)^{-1}</math>.
			\|}

Vipul: /* Proof */

2014-12-19T16:17:16Z

Proof

← Older revision		Revision as of 16:17, 19 December 2014
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	==Proof==		==Proof==

	~~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>.		'''Given''': <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>.

	~~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>.		'''To prove''': There exists <math>x \in G</math> such that <math>\sigma(H) = xHx^{-1}</math>

			'''Proof''': 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>, so setting <math>x = g^{-1}h</math> works.

Vipul at 16:15, 19 December 2014

2014-12-19T16:15:11Z

← Older revision		Revision as of 16:15, 19 December 2014
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	{{subgroup metaproperty satisfaction}}		{{subgroup metaproperty satisfaction\|
			property = automorph-conjugate subgroup\|
			metaproperty = transitive subgroup property}}

	==Statement==		==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]].~~

	~~===Symbolic statement===~~

	~~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>.

	==Proof==		==Proof==

Vipul: 2 revisions

2008-05-07T22:51:03Z

2 revisions

← Older revision	Revision as of 22:51, 7 May 2008
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