Given a group G acting both on the left and the right of a set X, say the actions are compatible if $g \cdot (x \cdot g') = (g\cdot x)\cdot g'$ for all $g,g'\in G$ and $x\in X$.

Here are some examples of compatible $G$-actions:

1) For any left action (resp. right action), that action and the trivial right action (resp left action) on the same set are compatible.

2) For any group G, its actions on itself by left and right multiplication are compatible (this is just associativity of the group law).

3) For any commutative group G, take any left or right action on a set X and let it act by the same map to Aut(X) on the other side, and by commutativity these actions are compatible.

We can also compose these examples with products and surjective group maps to obtain more compatible actions.

My question is, are there other naturally arising compatible actions?

  • $\begingroup$ I had voted to close earlier because I felt the question was too elementary for this site. $\endgroup$ – Todd Trimble Sep 4 '12 at 0:37

Suppose we have a group $G$ acting on the right and on the left as above on a set $X$. These actions give rise to a left action of $G\times G$ on $X$ by setting $(g_1,g_2)\cdot x=g_1\cdot x \cdot g_2^{-1}$. Conversely, if we have a left action of $G\times G$ on a set $X$, we can construct compatible left and right actions by setting $g\cdot x= (g,e)\cdot x$ and $x\cdot g= (e,g^{-1})\cdot x$. The compatibility follows from the fact that the factors of $G\times G$ commute.

Of course, instead of left actions of $G\times G$ one can also consider right actions.

  • $\begingroup$ Sean -- welcome. Let me also note that this can be generalized a bit. Namely, suppose we have two groups $G$ and $H$ acting on the same set $X$, one on the left, the other on the right. Then one can define the compatibility between the two actions as above; compatible pairs of actions will be in bijection with (left or right) actions of $G\times H$ on $X$. $\endgroup$ – algori Sep 3 '12 at 19:40
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    $\begingroup$ Note that this is similar to the situation for bimodules over a ring. If $A$ is a ring, then an $A-A$ bimodule is the same thing as an $A \otimes A^{op}$-module. $\endgroup$ – MTS Sep 3 '12 at 21:24

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