= Solution
Let $N\mathrel{\trianglelefteq}G$ be proper. If $7$ divides $|N|$, then $N$ contains a Sylow $7$-subgroup of $G$, and normality makes it contain all eight of them. Thus $n_7(N)=8$, so the Sylow divisibility theorem gives
$$
56=7\cdot8\mid |N|.
$$
Since $|N|$ divides 168 and $N$ is proper, this forces $|N|=56$.
Every nontrivial normal <subgroup> is transitive by <normal subgroup orbits in a faithful prime-degree action>. Hence
$$
|N\cap G_x|=|N_x|=56/7=8.
$$
But $N\cap G_x$ is normal in $G_x\cong S_4$, whereas $S_4$ has three, rather than one, Sylow $2$-subgroups. This contradiction proves that no proper normal <subgroup> has order divisible by seven.
If instead $3$ divides $|N|$, normality makes $N$ contain all 28 Sylow $3$-subgroups of $G$. Hence $n_3(N)=28$, so $28$ divides $|N|/3$ and therefore $84$ divides $|N|$. A proper such <subgroup> would have order 84, which is divisible by seven and has just been ruled out.
Finally, if $N$ is any nontrivial normal <subgroup>, the prime-degree orbit argument makes $N$ transitive, so seven divides $|N|$. This is impossible for a proper normal <subgroup>. Therefore $G$ is simple.
Back to article page
