$\frac{1}{n}$ is the exact threshold probability for a random graph $\Gamma(n,p)$ to be planar As far as I know, $\hat{p} = \frac{1}{n}$ is the exact threshold probability for a random graph $\Gamma(n,p)$ to be planar, i.e. if $p = p(n) \geq (1 + \varepsilon) \hat{p}$ for some $\varepsilon > 0$, then $\mathbb{P}(\Gamma(n,p) \text{ is planar}) \to 0$ as $n \to \infty$ and if $p = p(n) \leq (1 - \varepsilon) \hat{p}$ for some $\varepsilon > 0$, then $\mathbb{P}(\Gamma(n,p) \text{ is planar}) \to 1$ as $n \to \infty$. But I don't fully understand how to accurately prove it.
Suppose $p = p(n) \leq (1 - \varepsilon) \hat{p}$. As long as planarity is a decreasing property, it is enough to prove $\mathbb{P}(\Gamma(n,p) \text{ is planar}) \to 1$ for the case, when $pn \to (1-\varepsilon) < 0$. It is known that if $pn \to (1-\varepsilon) < 0$, then the probability of the components of $\Gamma(n,p)$ being only trees and unicyclic graphs tends to $1$. But trees and unicyclic graphs are planar, so  $\mathbb{P}(\Gamma(n,p) \text{ is planar}) \to 1$. Is this argument correct?
Suppose now that $p = p(n) \geq (1 + \varepsilon) \hat{p}$. In this case, as far as I know, for any $r>0$ the probability of $\Gamma(n,p)$ containing a complete graph with $r$ vertices ($K_r$) tends to $1$. I don't know, how to prove it, but if it is true, then $\mathbb{P}(\Gamma(n,p) \text{ is planar}) \to 0$, as long as $K_5$ is not planar. Is there a simpler way to prove it? As far as I know, in this case if we assume additionally that $np \to (1+\varepsilon) > 0$ then with the probability tending to $1$ the size of the largest component of $\Gamma(n,p)$, devided by $n$, tends to $\beta$ in probability, where $\beta$ is the solution of the equation $\beta + e^{-(1 + \varepsilon)\beta} = 1$. Is this statement somehow applicable to my question?
Any help would be appreciated.
 A: Your argument below the threshold is correct.
Above the threshold, it is false that you get $K_r$ for all $r$ once $p \ge \frac{1+\varepsilon}{n}$, and in particular it is false that you get $K_5$ at that point. Since $K_5$ has $5$ vertices and $10$ edges, the expected number of $K_5$ subgraphs is $O(n^5 p^{10})$, which tends to $0$ when $p \ll \frac1{\sqrt n}$. By a more complicated argument, that is the actual threshold. Regardless, when $p = \frac{1+\varepsilon}{n}$, you have a constant number of copies of $K_3$, but no copies of $K_4$ or $K_5$.
On the other hand, looking for a subdivision of $K_5$ or $K_{3,3}$ in the random graph sounds messy but doable. There are many approaches.
In particular, we obtain a subdivision of $K_{3,3}$ if we find a cycle containing vertices $v_1, v_2, v_3, v_4, v_5, v_6$ in that order with edges $v_1v_4, v_2v_5, v_3v_6$. (This is not a fully general form of a subdivision, but it should be good enough.) Once $p \ge \frac{1+\varepsilon}{n}$, we expect to find cycles of linear length, which gives us many choices of $v_1, \dots, v_6$, so completing the subdivision should be easy. You could do a "sprinkling" argument: hold some small fraction of your edges in reserve, and show that with high probability the reserve will complete the subdivision.
(If you are unfamiliar with such arguments, I suggest the chapter on the Erdős-Rényi phase transition from Alon and Spencer's Probabilistic Method. This chapter has many examples showing how to carefully deal with random graphs when $p$ is near $\frac1n$.)
