I am trying to work through an example of a Hamiltonian action, but I want to check my work so far on a lemma on Lie groups.
Let $G$ and $H$ be Lie groups, and suppose $G$ acts on $H$ via automorphisms. Then we form the semidirect product $G \ltimes H$ with $(g_1,h_1)(g_2,h_2) = (g_1g_2,h_1 (g_1 \cdot h_2))$. As a manifold, $G \ltimes H$ is diffeomorphic to the product $G \times H$, but the group structure is obviously different than the direct product of groups. Likewise, we can expect the Lie algebra to be isomorphic to $\mathfrak{g} \oplus \mathfrak{h}$ as vector spaces, but with different brackets.
The group action induces a Lie algebra homomorphism $\Psi: \mathfrak{g} \to \text{der } \mathfrak{h}$ by taking the differential of the restricted conjugacy maps $\phi_g(h) = ghg^{-1}$, where multiplication is done in $G \ltimes H$, viewing both $G$ and $H$ as subgroups. In fact, $\Psi$ is the adjoint representation on $G \ltimes H$ which comes inherent with the Lie group structure, without making reference to the Lie bracket we are trying to discover.
According to Wikipedia, the Lie bracket is given by $[(X_1,Y_1),(X_2,Y_2)] = [Y_1,Y_2]_\mathfrak{h} + \text{ad}_{X_1}Y_2 - \text{ad}_{X_2}Y_1 + [X_1,X_2]_\mathfrak{g}$.
Now let's turn to the concrete example I am trying to consider, the group of distance preserving maps on $\mathbb{R}^n$, denoted $E(n)$, the $n$th Euclidean group. It is known that $E(n) = O(n) \ltimes \mathbb{R}^n$, and $\mathfrak{o}(n)$ is the space of $n \times n$ skew symmetric matrices. The Lie algebra, as a vector space, is therefore the direct sum $\{ n \times n \text{ skew symmetric matrices } \} \oplus \mathbb{R}^n$.
According to this Stack Exchange question, the Lie algebra $\mathfrak{e}(n)$ can be encoded as the set of matrices $\left\{\begin{bmatrix}0 & 0^t \\ \mathbf{a} & X \end{bmatrix}:X \in \mathfrak{o}(n),\mathbf{a} \in \mathbb{R}^n \right\}$. Then will we have that the matrix commutator will correspond to the Lie bracket for the semidirect product?