# Left regular action isomorphic to the right regular action

I am reading something on regular groups and I have a question on why the left and right regular actions are isomorphic.

Let $G$ be a group. Consider the homomorphism $\rho: G \rightarrow Sym(G), g \mapsto \rho_g$, where $\rho_g: x \mapsto xg$. Recall that the image $R(G)$ of this homomorphism is isomorphic to $G$ by Cayley's theorem. Similarly, the left regular representation $L(G):=\{\lambda_g: g \in G\}$, where $\lambda_g: x \mapsto g^{-1}x$, is also isomorphic to $G$, whence $L(G) \cong R(G)$ since isomorphism is a transitive relation. Alternatively, $\rho_g \mapsto \lambda_g$ gives a direct isomorphism between $R(G)$ and $L(G)$.

In the text I'm reading, action by right multiplication is defined as $\mu(x,g) = xg$. And then the text says

There is also a left regular representation of $G$ on itself, given by the rule that $\mu(x,g) = g^{-1} x$. (The inverse is required here to make a proper action: note that $h^{-1}(g^{-1}x)=(gh)^{-1}x$.) It is, as the name says, also a regular action, and so must be isomorphic to the right regular action: indeed the map $x \mapsto x^{-1}$ is an isomorphism.''

My question is on the last part that says that $x \mapsto x^{-1}$ is an isomorphism. I thought that this was only true for abelian groups.

Perhaps the fact that $x \mapsto x^{-1}$ is a bijection is what is required here, given that the text defines two actions to be isomorphic as follows: If $\rho: \Omega \times G \rightarrow \Omega$ and $\lambda: \Delta \times G \rightarrow \Delta$ are $G$-actions, then they are said to be isomorphic if there is a bijection $\theta: \Omega \rightarrow \Delta$ such that $\forall \alpha \in \Omega, g \in G$, we have $\rho(\alpha,g) \theta = \lambda(\alpha \theta,g)$, i.e., each element of $G$ effects essentially the same permutation on these two sets. It can be verified that this definition of isomorphic actions is satisfied for the left and right regular actions using the correspondence $\theta:x \rightarrow x^{-1}$.

So my guess is that the author meant to say the inversion map is a bijection rather than an isomorphism. I couldn't think of a way to prove that $L(G) \cong R(G)$ or to prove that the left and right regular actions are isomorphic using the fact that $x \mapsto x^{-1}$ is a homomorphism (in fact, I think this inversion map is a homomorphism iff the group is abelian).

$x \mapsto x^{-1}$ is an isomorphism of $G$-sets, not an isomorphism of groups. (A $G$-set is a set equipped with an action of $G$.)