I am reading an article in which I do not understand some equivariance property about the momentum map. Let $G$ be a Lie group acting on a manifold $Q$. The action is denoted $(g,q) \, \mapsto \, q \cdot g$. The standard notation $T_{id_{G}}G = \mathfrak{g}$ is used to represent the Lie algebra of $G$.
They define the momentum map by : $(q,p) \in Q \times T^*Q \, \mapsto \, m(q,p) \in \mathfrak{g}^*$ with :
$$ (m(q,p) \vert \xi) = (p \vert q \cdot \xi) , \, \xi \in \mathfrak{g} $$
Considering, $g \in G$, $q \in Q$, $p \in T^*Q$ and $\xi \in \mathfrak{g}$, they write the following equalities :
\begin{eqnarray*} (m(q \cdot g, p \cdot g) \vert \xi) & = & (p \cdot g \vert (q \cdot g) \cdot \xi) \\ & = & (p \cdot g \vert q \cdot (g\xi)) \\ & = & (p \vert (q \cdot (g\xi)) \cdot g^{-1}) \\ & = & (p \vert q \cdot (g\xi)g^{-1}) \end{eqnarray*}
where $g\xi$ denotes the derivative of $h \, \mapsto \, gh$ in $h = id_{G}$ along the direction $\xi$ and $(g\xi)g^{-1}$ the derivative of $h \, \mapsto \, hg^{-1}$ in $h=g$ along the direction $g\xi$.
The first equality is just the definition of the momentum map. I do not understand the second equality. If I consider the left translation $L_{g} \, : G \, \rightarrow \, G ; h \, \mapsto \, gh$, I have $D_{id_{G}} L_{g} \, : \, \mathfrak{g} \, \rightarrow \, \mathfrak{g}$ the differential of $L_{g}$ at $id_{G}$. With their notations, is $g\xi$ equal to $D_{id_{G}} L_{g} \cdot \xi$ ?
Thank you for your help.