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$G$ a group, $A$ a $G$-group ($G\to \operatorname{Aut} A$ with $\sigma\mapsto(a\mapsto^{\sigma}a)$)

$X$ be a set. $X$ with an action of $G$ and $X$ with an action of $A$ s.t. the three actions ($G$ on $A$, $G$ on $X$ and $A$ on $X$) are compatible ($\forall$ $\sigma\in G$, $a\in A$, $x\in X$: $\sigma(a x)=(^{\sigma}a) (\sigma x))$

$\Leftrightarrow$

$X$ with an action of $A\rtimes G$ (formed with respect to $G$-group structure on $A$).

($^{\sigma}a = \sigma a \sigma^{-1}$)

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1 Answer 1

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Suppose we are given an action $\rho \colon (A \rtimes G) \times X \to X$. We define an $A$-Action $\rho_A\colon A \times X \to X$ by \[ \rho_A(a, x) := \rho\bigl((a,1), x\bigr) \] and a $G$-action by \[ \rho_G(g,x) := \rho\bigl((1,g), x\bigr) \] (Note that as $(1,g)(1,h) = (1,gh)$ and $(a,1)(b,1) = (ab,1)$ by definition of the semidirect product, so $\rho_G$ and $\rho_A$ are actions). We have for $a \in A$, $g \in G$: \begin{align*} \rho_G\bigl(g,\rho_A(a,x)\bigr) &= \rho\Bigl((1,g), \rho\bigl((a,1), x\bigr)\Bigr) \\&= \rho\bigl((1,g)(a,1), x\bigr) \\&= \rho\bigl((\sideset{^g}{}a,g), x\bigr)\\ \\&= \rho\bigl((\sideset{^g}{}a,1)(1,g), x\bigr)\\ \\&= \rho_A\bigl(\sideset{^g}{}a,\rho_G(g, x)\bigr) \end{align*} as wished.

The other way, given compatible actions $\rho_A\colon A \times X \to X$ and $\rho_G\colon G \times X \to X$, define $\rho \colon (A \rtimes G) \times X \to X$ by \[ \rho\bigl((a,g), x\bigr) = \rho_A\bigl(a, \rho_G(g,x)\bigr) \] We will show, that this is an action, so let $(a,g), (b,h) \in A \rtimes G$, we have using the compability \begin{align*} \rho\bigl((a,g), \rho((b,h), x)\bigr) &= \rho_A\biggl(a,\rho_G\Bigl(g, \rho_A\bigl(b,\rho_G(h,x)\bigr)\Bigr)\biggr)\\ &= \rho_A\biggl(a, \rho_A\Bigl(\sideset{^g}{}b, \rho_G\bigl(g, \rho_G(h, x)\bigr)\Bigr)\biggr)\\ &= \rho_A\bigl(a\sideset{^g}{}b, \rho_G(gh, x)\bigr)\\ &= \rho\bigl((a\sideset{^g}{}b, gh), x\bigr)\\ &= \rho\bigl((a,g)(b,h), x\bigr) \end{align*} As $(a,g)(b,h) = (a\sideset{^g}{}b, gh)$ in $A \rtimes G$.

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@Martha: Great answer so (+1). –  Nancy Rutkowskie Oct 15 '12 at 15:44

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