CTL and LTL logic difference I am trying to understand these subtle differences between LTL and CTL logic and one thing I simply don't have an idea how to understand.

Formula 

AG AF p 


in CTL and a formula

GF p


in LTL. Why aren't they equivalent? Can you give me an example, please? I tried finding and drawing things, but I simply can't find a counter example and there is no explanation in the lecture materials I have.
There are already some rather good related answers regarding LTL versus CTL. In a nutshell, LTL is first and foremost a logic of traces, and an LTL formula is true for a transition system $S$ if and only if it is true for each trace of $S$. CTL, on the other hand, is a branching-time logic, which can in a sense talk about multiple paths at the same time. 

One standard example here (not the one you give, about which more below) is a labelled transition system $S=(Q,T,q_0,L)$ with set of locations $Q=\{q_0,q_1,q_2\}$, set of transitions $T=\{(q_0,q_0),(q_0,q_1),(q_1,q_2),(q_2,q_2)\}$, and labelling given by $L(q_0)=L(q_2)=\{p\}$, $L(q_1)=\emptyset$. This system satisfies $FGp$, but not $AFAGp$, which can be seen as follows.

$FGp$ means that for every path $\pi=s_1,s_2,\dots$ in a given system, there is some point after which $p$ is always satisfied, i.e. there is some $i$ such that for all $j\geq i$, $p\in L(s_i)$. This is satisfied by $S$ since every path in $S$ either remains in $q_0$ forever (such that $p$ is always satisfied) or eventually gets to $p_2$ (after which $p$ is always satisfied). 

On the other hand, $AFAGp$ means that every path $\pi=s_1,s_2,\dots$ eventually reaches a state satisfying $AGp$, i.e. a state such that on every path $\pi'$ starting there, $p$ is always satisfied. Formally, this means that there is an $i$ such that for all $\pi'=s_1',s_2',\dots$ with $s_1'=s_i$ and all $j$, we have $p\in L(s_j')$. But in $S$, for the path which always remains in $q_0$, the transition to $q_1$, where $p$ is not satisfied, is always available, so that at no point of that path $AGp$ holds; therefore $AFAGp$ is not satisfied by $S$. 

As for your example, this is actually a case where two formulas are equivalent. This does happen. The proof is a little involved, but I can add it if you are interested.