There are no simple groups of order 240 I'm trying to prove that there are no simple groups of order $240$. So let $G$ be a simple group such that $|G|=240=2^4\cdot3\cdot5$. Then
$$n_2\in\{1,3,5,15\}\quad n_3\in\{1,4,10,40\}\quad n_5\in\{1,6,16\}$$
After some reasoning, we get $n_2=15$ and two mutually exclusive cases:


* If for every two distinct $P,Q\in\text{Sylow}_2(G)$ we have $P\cap Q=1$, then, by counting elements of order a power of $2$ we get only one $3$-Sylow and only one $5$-Sylow (so $G$ is not simple)

* If there are tow distinct $P,Q\in\text{Sylow}_2(G)$ such that $P\cap Q$ is not trivial then $|P\cap Q|\in\{2,4,8\}$ and I don't know how to proceed from here. Trying to prove $P\cap Q\unlhd G$ fails

 A: You may find $n_5\ne 6$ and $n_3\ne 4$ probably by the similar proof of $n_2 = 15$: $G$ cannot be embedded to $S_4$ and $S_6$. Moreover, $n_3\ne 40$ (see the answer here).
Therefore, we have $n_2 = 15$, $n_3 = 10$ and $n_5 = 16$. We use $N/C$ lemma to prove this is impossible:
Consider $G_5\in\mathrm{Syl}_5(G)$, which is isomorphic to the cyclic group $\mathbb{Z}_5$. Now $N_G(G_5)/C_G(G_5)$ is isomorphic to a subgroup of $\mathrm{Aut}(G_5)\cong\mathbb{Z}_4$, and $N_G(G_5)$ has order $\frac{240}{16} = 15$ (see here). So $C_G(G_5)$ has order $15$, and so $C_G(G_5)\cong\mathbb{Z}_{15}\cong\mathbb{Z}_5\times\mathbb{Z}_3$. It follow that $G_5$ is commutative with some $G_3\in\mathrm{Syl}_3(G)$.
Consider $N_G(G_3)$, which has order $\frac{240}{10} = 24$, and so $|C_G(G_3)| = 12$ or $24$. But $5\nmid 12$ and $5\nmid 24$, a contradiction to $G_5$ and $G_3$ are commutative.
Thus, $n_2 = 15$, $n_3 = 10$ and $n_5 = 16$ is impossible for $G$. So $G$ is not simple.
A: There is a mistake, or at least a gap, in the provided answer. The case
$n_3 = 16$ is not considered.  In addition, the case $n_3 = 40$ doesn't seem to be dealt with in the given link.  Here is a proof that there are no simple groups of order $240$.
Let $G$ be a simple group of order $240$.
We have $n_5 \in \{1,6,16\}$. Since $G$ is simple, we have $n_5 \ne 1$, and
$n_5 \ne 6$ because otherwise $G$ would be isomorphic to a subgroup of
$A_6$ but $240 \nmid 360$. Thus $n_5 = 16$ and so $G$ is isomorphic to a
subgroup of $A_{16}$.
Let $P_1, \ldots, P_{16}$ be the Sylow $5$-subgroups of $G$. Then
$|N_G(P_i)| = 15$, $1 \le i \le 16$.
Every group of order $15$ is a cyclic group and thus each
$N_G(P_i)$ contains a unique subgroup of order $5$, namely $P_i$.
It follows that each subgroup in $G$ of order $5$ lies in exactly one subgroup
$N_G(P_i)$.
Let $N_G(P_i) = \langle g_i \rangle$, where $g_i$ has order $15$.
Then $g_i^3$ has order $5$ and fixes $P_i$, but fixes no other Sylow $5$-subgroup.
Thus $g_i$ fixes $P_i$ but fixes no other Sylow $5$-subgroup $P_j$, $i \ne j$.
Therefore, $g_i$ is a $15$-cycle because the only element of
$A_{16}$ that fixes exactly one element and has order $15$ is a $15$-cycle.
It follows that $|N_G(P_i) \cap N_G(P_j)| = 1$ for all $i \ne j$ because the non-identity elements of $N_G(P_i)$ fix $P_i$ and move every other $P_k$.
This gives $16 \times 14 = 224$ distinct elements of orders $3, 5, 15$ in $G$.
This leaves only $16$ elements not accounted for, and thus there must be a
unique Sylow $2$-subgroup of $G$, a contradiction because $G$ is simple.
