I want to prove that $\lvert G\rvert=24$, then it has a non-trivial normal subgroup.
Here is my attempt:
$n_2$: number of 2-Sylow subgroup; $n_3$: number of 3-Sylow subgroup.
$n_2\in\{1,3\},\:\:n_3\in\{1,4\}$ by Sylow's Theorem.
Suposse $n_2=3$, $n_3=4$.
$n_3=4$ means there are 4 Sylow subgroups of order 3, which means 8 elements of order 3.
$n_2=3$ means there are 3 Sylow subgroups of order 8. Suppose $P_1,P_2,P_3$. Using $\lvert P_1P_2\rvert\leq 24$, then $2<\lvert P_1\cap P_2\rvert <8$, so it is 4.
Suppose $P_1=\{1,a_1,\ldots,a_7\}$.
$\lvert P_1\cap P_2\rvert=4 \Longrightarrow P_2=\{1,a_1,a_2,a_3,b_1,b_2,b_3,b_4\}$ for example.
$\lvert P_1\cap P_3\rvert=4$ and $\lvert P_3\cap P_2\rvert=4$ $\Longrightarrow P_3=\{1,a_7,a_6,a_5,b_1,b_2,b_3,c\}$ for example.
Then we have 12 different elements of even order.
Adding we get 21 different elements.