Skew group ring

Remarks

The product is associative because $\alpha_{gh} = \alpha_g \circ \alpha_h$: both $((rg)(sh))(tk)$ and $(rg)((sh)(tk))$ equal $r \alpha_g(s) \alpha_{gh}(t) \, ghk$. The identity is $1 \cdot e$, where $e$ is the identity of $G$, and $R$ is a subring through $r \mapsto r e$. If $\alpha$ is trivial, $R * G$ is the group ring $R[G]$. Example: let $G = \set{e, c}$ act on $\C$ by complex conjugation, $\alpha_c(z) = \bar z$. Then $R * G = \C * G$ has dimension $4$ over $\R$. The map sending $z$ to multiplication by $z$ and $c$ to conjugation, both as $\R$-linear maps of $\C = \R^2$, is an isomorphism onto the matrix ring $M_2(\R)$. Non-example: if $\alpha \colon G \to \Aut(R)$ is a map that is not a homomorphism, the same formula fails associativity, since $(1g)((1h)(s e)) = \alpha_g(\alpha_h(s)) gh$ while $((1g)(1h))(s e) = \alpha_{gh}(s) gh$. When a twisting cocycle is also allowed, $g h$ is replaced by a scalar multiple of $gh$, and one gets a crossed product.