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Let $A$, $B$ be subgroups of $G$ and $AB=\{ab:a\in A, b\in B\}\subseteq G$.
Show that the direct product $A\times B$ acts on the set $AB$ by $(a,b)x=axb^{-1}$ is a group action.
Show that there is only one orbit.

Let $a_1,a_2\in A$, $b_1,b_2$ and $x\in AB$ and show the axioms of a group action hold as follows $$ (a_1,b_1)(a_2,b_2)x=(a_1,b_1)a_2xb_2^{-1}=a_1a_2xb_2^{-1}b_1^{-1}=(a_1a_2)x(b_2^{-1}b_1^{-1})=(a_1a_2,b_1b_2)x=[(a_1,b_1)\times (a_2,b_2)]x $$ Showing the identity axiom for group actions is trivial.

The above seems correct but I'm stuck on how to show there is only one orbit. I've been thinking about how to show the stabilizer is limited to the identity but am not getting anywhere.

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To show that there is only one orbit, it suffices to show that the orbit of the identity element is the entire set $AB$. Any element of $AB$ is of the form $ab$ with $a\in A$ and $b\in B$, so we need to find $(a',b')\in A\times B$ such that $$(a',b')e=ab,$$ where $e$ denotes the identity element of $G$. Taking $(a',b')=(a,b^{-1})$ does the trick.

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