lie bracket is vanish iff translation is commute

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I want to prove following claim:

Given two vector fields $X$ and $Y$ on smooth manifold $M$, $[X,Y]=0$ if and only if $\Phi^X_t \circ \Phi^Y_s \circ \Phi^X_{-t} \circ \Phi^Y_{-s}(q)=q$ for $\forall q\in M$ and $\forall s,t\in \mathbb{R}$.

I tried to prove this by changing the form of lie bracket, $$[X,Y]_p(f) = \frac{\partial^2 H}{\partial s \partial t}(0,0)$$ where $H(s,t)=f\circ \Phi^X_t \circ \Phi^Y_s \circ \Phi^X_{-t}(p)$

However, it is difficult to extend the data at $(0,0)$ into whole $(s,t)$.

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Another approach to schow this is to consider $H(s)(t)=\Phi_{s}^X\circ\Phi_{t}^Y\circ\Phi^X_{-s}$ where $t_0$ is fixed, ${d\over{dt}}H(s)=(\Phi_s^X)^*(Y)$ and ${d\over{ds}}(\Phi_s^X)^*(Y)=(\Phi_s^X)^*[X,Y]$