Characteristicity is quotient-transitive - Revision history

Vipul: /* Proof */

2022-05-01T04:40:10Z

Proof

← Older revision		Revision as of 04:40, 1 May 2022
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	==Proof==		==Proof==

	'''Given''': A group <math>G</math>, subgroups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>G</math>, and <math>K/H</math> is characteristic in <math>G/H</math>. An automorphism <math>\sigma</math> of <math>G</math>		'''Given''': A group <math>G</math>, subgroups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>G</math>, and <math>K/H</math> is characteristic in <math>G/H</math>. An automorphism <math>\sigma</math> of <math>G</math>.

	'''To prove''': <math>\sigma(K) \subseteq K</math>		'''To prove''': <math>\sigma(K) \subseteq K</math>

Vipul: /* Proof */

2022-05-01T04:38:24Z

Proof

← Older revision		Revision as of 04:38, 1 May 2022
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	==Proof==		==Proof==

	'''Given''': A group <math>G</math>, subgroups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>G</math>, and <math>K/H</math> is characteristic in <math>G/H</math>		'''Given''': A group <math>G</math>, subgroups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>G</math>, and <math>K/H</math> is characteristic in <math>G/H</math>. An automorphism <math>\sigma</math> of <math>G</math>

	'''To prove''': <math>K~~</math> is characteristic in <math>G~~</math>		'''To prove''': <math>\sigma(K) \subseteq K</math>

	'''Proof''': ~~We pick any automorphism <math>\sigma</math> of <math>G</math>, and want to show that <math>\sigma(K) = K</math>. For this, first~~ observe that <math>\sigma(H) = H</math>, so <math>\sigma</math> induces an automorphism on the quotient <math>G/H</math>, by the rule <math>gH \mapsto \sigma(g)H</math>. Call this automorphism <math>\sigma'</math>.		'''Proof''': First observe that <math>\sigma(H) = H</math>, so <math>\sigma</math> induces an automorphism on the quotient <math>G/H</math>, by the rule <math>gH \mapsto \sigma(g)H</math>. Call this automorphism <math>\sigma'</math>.

	Then, <math>\sigma'</math> is an automorphism of <math>G/H</math>. Since <math>K/H</math> is characteristic in <math>G/H</math>, <math>\sigma'(K/H) = K/H</math>. Thus, for any <math>g \in K</math>, <math>\sigma'(gH) \in K/H</math>, and hence, unwrapping the definition, <math>\sigma(g) \in K</math>. Thus, <math>\sigma(K) \~~subset K</math>. Since the same holds for <math>\sigma^{-1}</math>, we conclude that <math>\sigma(K) =~~ K</math>, completing the proof.		Then, <math>\sigma'</math> is an automorphism of <math>G/H</math>. Since <math>K/H</math> is characteristic in <math>G/H</math>, <math>\sigma'(K/H) = K/H</math>. Thus, for any <math>g \in K</math>, <math>\sigma'(gH) \in K/H</math>, and hence, unwrapping the definition, <math>\sigma(g) \in K</math>. Thus, <math>\sigma(K) \subseteq K</math>, completing the proof.

Vipul at 05:41, 30 July 2013

2013-07-30T05:41:05Z

← Older revision		Revision as of 05:41, 30 July 2013
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	Suppose <math>H \le K \le G</math> are subgroups such that <math>H</math> is a [[characteristic subgroup]] of <math>G</math>, and <math>K/H</math> is a characteristic subgroup of <math>G/H</math>. Then, <math>K</math> is a characteristic subgroup of <math>G</math>.		Suppose <math>H \le K \le G</math> are subgroups such that <math>H</math> is a [[characteristic subgroup]] of <math>G</math>, and <math>K/H</math> is a characteristic subgroup of <math>G/H</math>. Then, <math>K</math> is a characteristic subgroup of <math>G</math>.

			==Related facts==

			===Similar facts===

			* [[Normality is quotient-transitive]]
			* [[Full invariance is quotient-transitive]]
			* [[Strict characteristicity is quotient-transitive]]

	==Proof==		==Proof==

Vipul at 05:05, 30 July 2013

2013-07-30T05:05:10Z

← Older revision		Revision as of 05:05, 30 July 2013
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	{{subgroup metaproperty satisfaction}}		{{subgroup metaproperty satisfaction\|
			property = characteristic subgroup\|
			metaproperty = quotient-transitive subgroup}}
			[[Difficulty level::1\| ]]
	==Statement==		==Statement==

Vipul: New page: {{subgroup metaproperty satisfaction}} ==Statement== ===Property-theoretic statement=== The subgroup property of being a characteristic subgroup satisfies the [[subgroup metapro...

2008-07-13T17:03:20Z

New page: {{subgroup metaproperty satisfaction}} ==Statement== ===Property-theoretic statement=== The subgroup property of being a characteristic subgroup satisfies the [[subgroup metapro...

New page

{{subgroup metaproperty satisfaction}}

==Statement==

===Property-theoretic statement===

The [[subgroup property]] of being a [[characteristic subgroup]] satisfies the [[subgroup metaproperty]] of being [[quotient-transitive subgroup property|quotient-transitive]].

===Statement with symbols===

Suppose <math>H \le K \le G</math> are subgroups such that <math>H</math> is a [[characteristic subgroup]] of <math>G</math>, and <math>K/H</math> is a characteristic subgroup of <math>G/H</math>. Then, <math>K</math> is a characteristic subgroup of <math>G</math>.

==Proof==

'''Given''': A group <math>G</math>, subgroups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>G</math>, and <math>K/H</math> is characteristic in <math>G/H</math>

'''To prove''': <math>K</math> is characteristic in <math>G</math>

'''Proof''': We pick any automorphism <math>\sigma</math> of <math>G</math>, and want to show that <math>\sigma(K) = K</math>. For this, first observe that <math>\sigma(H) = H</math>, so <math>\sigma</math> induces an automorphism on the quotient <math>G/H</math>, by the rule <math>gH \mapsto \sigma(g)H</math>. Call this automorphism <math>\sigma'</math>.

Then, <math>\sigma'</math> is an automorphism of <math>G/H</math>. Since <math>K/H</math> is characteristic in <math>G/H</math>, <math>\sigma'(K/H) = K/H</math>. Thus, for any <math>g \in K</math>, <math>\sigma'(gH) \in K/H</math>, and hence, unwrapping the definition, <math>\sigma(g) \in K</math>. Thus, <math>\sigma(K) \subset K</math>. Since the same holds for <math>\sigma^{-1}</math>, we conclude that <math>\sigma(K) = K</math>, completing the proof.