I'm trying to prove that no group of order 160 is simple. The following is my approach.
Let $G$ be a group of order $160$. Note $160 = 2^55$. I can easily get that $n_2 = 1$ or $5$ and $n_5 = 1$ or $16$ where $n_2, n_5$ are the numbers of Sylow 2, 5-subgroups, resp. Suppose that $n_2 = 5$ and $n_5 = 16$. I want to get some contradiction with this assumption. Since $n_2 = 5$, let $P_1, P_2, \cdots, P_5$ be the distinct Sylow 2-subgroups. Since $|P_iP_j| =\frac{|P_i||P_j|}{|P_i \cap P_j|} = \frac{2^{10}}{|P_i \cap P_j|}$ and $P_iP_j \subset G$, $\frac{2^{10}}{|P_i \cap P_j|} \leq 2^55$. Hence $|P_i \cap P_j| \geq 7$. Also we know that $P_i \cap P_j$ is a subgroup of $P_i$, the order of $P_i \cap P_j$ divides 32. Thus $|P_i \cap P_j| = 8$ or $16$ for $i \neq j$. Suppose that there is $i \neq j$ s.t $|P_i \cap P_j| = 16$. Since $P_i \cap P_j$ is a subgroup of $P_i$ and $P_j$, and the indices of $P_i \cap P_j$ is 2 in each $P_i$ and $P_j$, $P_i \cap P_j$ is a normal subgroup of $P_i, P_j$. Then the normalizer $N(P_i \cap P_j)$ contains $P_i \cup P_j$ and $|P_i \cap P_j| = 16$. Hence $|N(P_i \cap P_j)| \geq 32 + 32 - 16 = 48$. Also $32 | |N(P_i \cap P_j)|$, $|N(P_i \cap P_j)|$ must be 160, whence $N(P_i \cap P_j) = G$. This shows that $P_i \cap P_j$ is a normal subgroup of $G$, so $G$ is not simple.
Next, suppose that for each $i \neq j$, $|P_i \cap P_j| = 8$. I have a trouble with this case. Since $n_5 = 16$, there is $65 = 1 + 16*4$ distinct elements in Sylow 5 subgroups. I want to show that if $|P_i \cap P_j|=8$ for each $i\neq j$, the total number of distinct elements in Sylow 2-subgroups exceed $95 = 160-65$. But I just manage to prove that this number exceed at least $32+24+17+10+3 = 86$ using set theorical method. How can I use the group theorical method to prove this one? Any hint please. I saw other proof in this site, but I cannot understand this part.