Compatible pair of actions - Revision history

Vipul: /* Definition with left action convention */

2013-09-13T22:38:27Z

Definition with left action convention

← Older revision		Revision as of 22:38, 13 September 2013
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	Here is an equivalent formulation of these two conditions that is more convenient:		Here is an equivalent formulation of these two conditions that is more convenient:

	* <math>~~\alpha(~~g_1)(\beta(h)g_2) = \beta(\alpha(g_1)h)(c_{g_1}(g_2)) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>c_{g_1^{-1}}g_2</math> of the preceding formulation).		* <math>c_{g_1}(\beta(h)g_2) = \beta(\alpha(g_1)h)(c_{g_1}(g_2)) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>c_{g_1^{-1}}g_2</math> of the preceding formulation).
	* <matH>~~\beta(~~h_1)(\alpha(g)h_2) = \alpha(\beta(h_1)g)(c_{h_1}(h_2)) \ \forall \ g \in G, h_1, h_2 \in H</math> (the <math>h_2</math> here is the <math>c_{h_1^{-1}}h_2</math> of the preceding formulation)		* <matH>c_{h_1}(\alpha(g)h_2) = \alpha(\beta(h_1)g)(c_{h_1}(h_2)) \ \forall \ g \in G, h_1, h_2 \in H</math> (the <math>h_2</math> here is the <math>c_{h_1^{-1}}h_2</math> of the preceding formulation)

	In the <matH>\cdot</math> notation, these become:		In the <matH>\cdot</math> notation, these become:

Vipul: /* Symmetry in the definition */

2013-07-19T19:44:32Z

Symmetry in the definition

← Older revision		Revision as of 19:44, 19 July 2013
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	==Symmetry in the definition==		==Symmetry in the definition==

	Given groups <math>G</math> and <math>H</math> with actions <math>\alpha:G \to \operatorname{Aut}(H)</math> and <math>\beta:H \to \operatorname{Aut}(G)<math>, <math>\alpha</math> is compatible with <math>\beta</math> if and only if <math>\beta</math> is compatible with <math>\alpha</math>. In other words, the definition of compatibility is symmetric under interchanging the roles of the two groups.		Given groups <math>G</math> and <math>H</math> with actions <math>\alpha:G \to \operatorname{Aut}(H)</math> and <math>\beta:H \to \operatorname{Aut}(G)</math>, <math>\alpha</math> is compatible with <math>\beta</math> if and only if <math>\beta</math> is compatible with <math>\alpha</math>. In other words, the definition of compatibility is symmetric under interchanging the roles of the two groups.

	==Particular cases==		==Particular cases==

Vipul: /* Definition with left action convention */

2013-07-19T19:44:15Z

Definition with left action convention

← Older revision		Revision as of 19:44, 19 July 2013
Line 15:		Line 15:
	Here is an equivalent formulation of these two conditions that is more convenient:		Here is an equivalent formulation of these two conditions that is more convenient:

	* <math>\alpha(g_1)(\beta(h)g_2) = \beta(\alpha(g_1)h)(c_{g_1}(g_2)) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>g_1^{-1} ~~\cdot~~ g_2</math> of the preceding formulation).		* <math>\alpha(g_1)(\beta(h)g_2) = \beta(\alpha(g_1)h)(c_{g_1}(g_2)) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>c_{g_1^{-1}}g_2</math> of the preceding formulation).
	* <matH>\beta(h_1)(\alpha(g)h_2) = \alpha(\beta(h_1)g)(c_{h_1}(h_2)) \ \forall \ g \in G, h_1, h_2 \in H</math> (the <math>h_2</math> here is the <math>h_1^{-1} ~~\cdot~~ h_2</math> of the preceding formulation)		* <matH>\beta(h_1)(\alpha(g)h_2) = \alpha(\beta(h_1)g)(c_{h_1}(h_2)) \ \forall \ g \in G, h_1, h_2 \in H</math> (the <math>h_2</math> here is the <math>c_{h_1^{-1}}h_2</math> of the preceding formulation)

	In the <matH>\cdot</math> notation, these become:		In the <matH>\cdot</math> notation, these become:

Vipul: /* Definition with left action convention */

2013-06-23T23:35:57Z

Definition with left action convention

← Older revision		Revision as of 23:35, 23 June 2013
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	Here is an equivalent formulation of these two conditions that is more convenient:		Here is an equivalent formulation of these two conditions that is more convenient:

			* <math>\alpha(g_1)(\beta(h)g_2) = \beta(\alpha(g_1)h)(c_{g_1}(g_2)) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>g_1^{-1} \cdot g_2</math> of the preceding formulation).
			* <matH>\beta(h_1)(\alpha(g)h_2) = \alpha(\beta(h_1)g)(c_{h_1}(h_2)) \ \forall \ g \in G, h_1, h_2 \in H</math> (the <math>h_2</math> here is the <math>h_1^{-1} \cdot h_2</math> of the preceding formulation)

			In the <matH>\cdot</math> notation, these become:

	* <math>g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>g_1^{-1} \cdot g_2</math> of the preceding formulation).		* <math>g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>g_1^{-1} \cdot g_2</math> of the preceding formulation).

Vipul: /* Particular cases */

2013-06-23T17:57:55Z

Particular cases

← Older revision		Revision as of 17:57, 23 June 2013
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	* [[Trivial pair of actions is compatible]]: If both the actions are trivial, i.e., both the homomorphisms <math>\alpha,\beta</math> are trivial maps, then they form a compatible pair.		* [[Trivial pair of actions is compatible]]: If both the actions are trivial, i.e., both the homomorphisms <math>\alpha,\beta</math> are trivial maps, then they form a compatible pair.
	* [[Compatible with trivial action iff image centralizes inner automorphisms]]: Suppose the homomorphism <math>\alpha</math> is trivial. In that case, the homomorphism <math>\beta</math> is compatible with <math>\alpha</math> if and only if the image of <math>\beta</math> in <math>\operatorname{Aut}(G)</math> is in the [[centralizer]] <math>C_{\operatorname{Aut}(G)}(\operatorname{Inn}(G))</math>.		* [[Compatible with trivial action iff image centralizes inner automorphisms]]: Suppose the homomorphism <math>\alpha</math> is trivial. In that case, the homomorphism <math>\beta</math> is compatible with <math>\alpha</math> if and only if the image of <math>\beta</math> in <math>\operatorname{Aut}(G)</math> is in the [[centralizer]] <math>C_{\operatorname{Aut}(G)}(\operatorname{Inn}(G))</math>.
	* If <math>G,H</matH> are both subgroups of some group <math>Q</math> that [[normalizer\|normalize]] each other (i.e., each is contained in the normalizer of the other), and <math>\alpha,\beta</math> are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in <math>Q</math> and checking the conditions simply amounts to checking two words to be equal.		* [[Conjugation actions between subgroups that normalize each other are compatible]]: If <math>G,H</matH> are both subgroups of some group <math>Q</math> that [[normalizer\|normalize]] each other (i.e., each is contained in the normalizer of the other), and <math>\alpha,\beta</math> are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in <math>Q</math> and checking the conditions simply amounts to checking two words to be equal.

Vipul at 17:56, 23 June 2013

2013-06-23T17:56:19Z

← Older revision		Revision as of 17:56, 23 June 2013
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	* <math>h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H</math> (the <math>h_2</math> here is the <math>h_1^{-1} \cdot h_2</math> of the preceding formulation)		* <math>h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H</math> (the <math>h_2</math> here is the <math>h_1^{-1} \cdot h_2</math> of the preceding formulation)

			===Definition with right action convention===

			We can give a corresponding definition using the right action convention, but the literature uses the left action convention, so this definition is intended purely as an illustrative exercise. {{fillin}}

			==Symmetry in the definition==

			Given groups <math>G</math> and <math>H</math> with actions <math>\alpha:G \to \operatorname{Aut}(H)</math> and <math>\beta:H \to \operatorname{Aut}(G)<math>, <math>\alpha</math> is compatible with <math>\beta</math> if and only if <math>\beta</math> is compatible with <math>\alpha</math>. In other words, the definition of compatibility is symmetric under interchanging the roles of the two groups.
	==Particular cases==		==Particular cases==

Vipul: /* Particular cases */

2013-06-23T17:38:53Z

Particular cases

← Older revision		Revision as of 17:38, 23 June 2013
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	* [[Trivial pair of actions is compatible]]: If both the actions are trivial, i.e., both the homomorphisms <math>\alpha,\beta</math> are trivial maps, then they form a compatible pair.		* [[Trivial pair of actions is compatible]]: If both the actions are trivial, i.e., both the homomorphisms <math>\alpha,\beta</math> are trivial maps, then they form a compatible pair.
	* [[Compatible with trivial action iff image centralizes inner automorphisms]]		* [[Compatible with trivial action iff image centralizes inner automorphisms]]: Suppose the homomorphism <math>\alpha</math> is trivial. In that case, the homomorphism <math>\beta</math> is compatible with <math>\alpha</math> if and only if the image of <math>\beta</math> in <math>\operatorname{Aut}(G)</math> is in the [[centralizer]] <math>C_{\operatorname{Aut}(G)}(\operatorname{Inn}(G))</math>.
	* If <math>G,H</matH> are both subgroups of some group <math>Q</math> that [[normalizer\|normalize]] each other (i.e., each is contained in the normalizer of the other), and <math>\alpha,\beta</math> are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in <math>Q</math> and checking the conditions simply amounts to checking two words to be equal.		* If <math>G,H</matH> are both subgroups of some group <math>Q</math> that [[normalizer\|normalize]] each other (i.e., each is contained in the normalizer of the other), and <math>\alpha,\beta</math> are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in <math>Q</math> and checking the conditions simply amounts to checking two words to be equal.

Vipul: /* Particular cases */

2013-06-23T17:37:02Z

Particular cases

← Older revision		Revision as of 17:37, 23 June 2013
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	* [[Trivial pair of actions is compatible]]: If both the actions are trivial, i.e., both the homomorphisms <math>\alpha,\beta</math> are trivial maps, then they form a compatible pair.		* [[Trivial pair of actions is compatible]]: If both the actions are trivial, i.e., both the homomorphisms <math>\alpha,\beta</math> are trivial maps, then they form a compatible pair.
			* [[Compatible with trivial action iff image centralizes inner automorphisms]]
	* If <math>G,H</matH> are both subgroups of some group <math>Q</math> that [[normalizer\|normalize]] each other (i.e., each is contained in the normalizer of the other), and <math>\alpha,\beta</math> are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in <math>Q</math> and checking the conditions simply amounts to checking two words to be equal.		* If <math>G,H</matH> are both subgroups of some group <math>Q</math> that [[normalizer\|normalize]] each other (i.e., each is contained in the normalizer of the other), and <math>\alpha,\beta</math> are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in <math>Q</math> and checking the conditions simply amounts to checking two words to be equal.

Vipul at 17:36, 23 June 2013

2013-06-23T17:36:20Z

← Older revision		Revision as of 17:36, 23 June 2013
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	* <math>g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>g_1^{-1} \cdot g_2</math> of the preceding formulation).		* <math>g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>g_1^{-1} \cdot g_2</math> of the preceding formulation).
	* <math>h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H</math> (the <math>h_2</math> here is the <math>h_1^{-1} \cdot h_2</math> of the preceding formulation)		* <math>h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H</math> (the <math>h_2</math> here is the <math>h_1^{-1} \cdot h_2</math> of the preceding formulation)

	~~==Facts==~~

	* [[Compatible with trivial action iff trivial action]]

	==Particular cases==		==Particular cases==

	* If both the actions are trivial, i.e., both the homomorphisms <math>\alpha,\beta</math> are trivial maps, then they form a compatible pair.		* [[Trivial pair of actions is compatible]]: If both the actions are trivial, i.e., both the homomorphisms <math>\alpha,\beta</math> are trivial maps, then they form a compatible pair.
	* If <math>G,H</matH> are both subgroups of some group <math>Q</math> that [[normalizer\|normalize]] each other (i.e., each is contained in the normalizer of the other), and <math>\alpha,\beta</math> are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in <math>Q</math> and checking the conditions simply amounts to checking two words to be equal.		* If <math>G,H</matH> are both subgroups of some group <math>Q</math> that [[normalizer\|normalize]] each other (i.e., each is contained in the normalizer of the other), and <math>\alpha,\beta</math> are the actions of the groups on each other by conjugation, then they form a compatible pair. Note that in this case, all the actions are just conjugation in <math>Q</math> and checking the conditions simply amounts to checking two words to be equal.

Vipul at 16:52, 23 June 2013

2013-06-23T16:52:38Z

← Older revision		Revision as of 16:52, 23 June 2013
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	* <math>g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>g_1^{-1} \cdot g_2</math> of the preceding formulation).		* <math>g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H</math> (the <math>g_2</math> here is the <math>g_1^{-1} \cdot g_2</math> of the preceding formulation).
	* <math>h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H</math> (the <math>h_2</math> here is the <math>h_1^{-1} \cdot h_2</math> of the preceding formulation)		* <math>h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H</math> (the <math>h_2</math> here is the <math>h_1^{-1} \cdot h_2</math> of the preceding formulation)

			==Facts==

			* [[Compatible with trivial action iff trivial action]]

	==Particular cases==		==Particular cases==