Compatible pair of actions: Difference between revisions

Revision as of 00:42, 10 June 2012

Definition

Definition with left action convention

Suppose $G$ and $H$ are groups. Suppose $α : G \to A u t (H)$ is a homomorphism of groups, defining a group action of $G$ on $H$ . Suppose $β : H \to A u t (G)$ is a homomorphism of groups, defining a group action of $H$ on $G$ . For $g \in G$ , denote by $c_{g} : G \to G$ the conjugation map by $g$ . See group acts as automorphisms by conjugation. Then, we say that the actions $α, β$ form a compatible pair if both these conditions hold:

$β (α (g_{1}) (h)) (g_{2}) = c_{g_{1}} (β (h) (c_{g_{1}^{- 1}} (g_{2})))) \forall g_{1}, g_{2} \in G, h \in H$
$α (β (h_{1}) (g)) (h_{2}) = c_{h_{1}} (α (g) (c_{h_{1}^{- 1}} (h_{2}))) \forall h_{1}, h_{2} \in H, g \in G$

The above expressions are easier to write down if we use $\cdot$ to denote all the actions. In that case, the conditions read:

$(g_{1} \cdot h) \cdot g_{2} = g_{1} \cdot (h \cdot (g_{1}^{- 1} \cdot g_{2})) \forall g_{1}, g_{2} \in G, h \in H$
$(h_{1} \cdot g) \cdot h_{2} = h_{1} \cdot (g \cdot (h_{1}^{- 1} \cdot h_{2})) \forall h_{1}, h_{2} \in H, g \in G$

Particular cases

If both the actions are trivial, i.e., both the homomorphisms $α, β$ are trivial maps, then they form a compatible pair.
If $G, H$ are both subgroups of some group $Q$ that normalize each other (i.e., each is contained in the normalizer of the other), and $α, β$ 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 $Q$ and checking the conditions simply amounts to checking two words to be equal.

Revision as of 00:41, 10 June 2012 (view source) Vipul (talk \| contribs) (→‎Particular cases) ← Older edit		Revision as of 00:42, 10 June 2012 (view source) Vipul (talk \| contribs) (→‎Particular cases) Newer edit →
Line 16:		Line 16:

	* 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 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.		* 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.