(This is the "standard" example which proves that CTL is not a superset of LTL, in terms of expressiveness)
Consider this system:
- state $q0$, transitions to $q0$ and $q1$
- state $q1$, transition to $q2$
- state $q2$, transition to $q2$
Let $p$ be true in $q0,q2$ but not in $q1$.
(p) (not p) (p)
q0 ----> q1 ----> q2
| ^ | ^
State $q0$ does not satisfy formula $AF[AGp]$. Take the infinite path $q0,q0,q0,\ldots$. There is no state in the path satisfying $AGp$ since we can only pick $q0$ and from there the path $q0,q1,q2,q2,q2,\ldots$ does not satisfy $Gp$ since $q1$ does not satisfy $p$.
State $q0$ however satisfies $A[FGp]$. Take any infinite path starting from $q0$. Either it is of the form $q0,q0,q0,\ldots$ or of the form $q0,q0,q0,\ldots,q0,q1,q2,q2,q2,\ldots$. Both forms satisfy $FGp$.