Two equivalent definitions of (strongly) Hamiltonian actions? Let $(M, \omega)$ be symplectic acted on by a Lie group $G$ and suppose $i_{\underline{X}} \omega$ is exact.
Notation:

*

*For $X \in \frak{g}$, $\underline{X}$ is the fundamental vector field on $M$ associated to $X$.

*$H_X \in \mathcal{C}^\infty(M)$ is the hamiltonian of $X \in \mathfrak{g}$ i.e. $i_{\underline{X}} \omega = d H_X$.

*The moment map $\mu:M \rightarrow \frak{g}^*$ is defined by $\mu(x)(X):= H_X(x)$.

I am used to defining (strongly) Hamiltonian actions with the following conditions, which I have seen in multiple texts:
$$ \omega(\underline{X}, \underline{Y}) = H_{[X,Y]} $$
for any $X,Y \in \frak{g}$, and the association $X \in \mathfrak{g} \mapsto H_X$ is $\mathbb{R}$-linear.
In the 1982 paper of Duistermaat and Heckman, "On the variation..." where they work with an abelian $G$, the condition reduces to:
$$ \omega(\underline{X}, \underline{Y}) = 0 $$
What they say is that this is equivalent to the fact that $G$ acts along the fibers of the moment map, which is another way to say:
$$ H_X(gx) = H_X(x)$$
for any $x \in M, g \in G, X \in \frak{g}$.
How would one show this?
 A: As commented, when $\mathfrak g$ is abelian and the action is strongly Hamiltonian in the general sense, we have
$$
\omega(\underline X,\underline Y)=H_{[X,Y]}=H_0=0
$$
thus the general notion of strongly Hamiltonian action reduces in this case to the condition $\omega(\underline X,\underline Y)=0$.
Next the derivative along $\underline Y$ of the function on $M$ given by $x\mapsto H_{X}(x)$ is
$$
\underline Y(H_{X})=\{H_{Y},H_{X}\}=0
$$
because all Hamiltonians are in involution.
Hence $H_{X}(x)$ in invariant under the infinitesimal action of $G$ and so
$$
H_{X}(gx)
$$
does not depend on $g\in G$.
EDIT: to address the comment, more details on the last statement. Any $g\in G$ can be written as $\exp(Y)$ for some $Y\in\mathfrak g$ (here we use that $G$ is connected and abelian). Therefore the function of $t\in[0,1]$ given by
$$
f(t):=H_X(\exp(t Y)x)
$$
(for any fixed $x\in M$) satisfies
$$
f(0)=H_X(x),\qquad f(1)=H_X(gx).
$$
However, $f(t)$ is constant because
$$
\frac{\mathrm d}{\mathrm d t}f(t) = \frac{\mathrm d }{\mathrm d t} H_X(\exp(t Y)x)=\{H_{Y},H_{X}\}(\exp(t Y)x)=0
$$
and $[0,1]$ is connected.
