A group of order $12p$, $p>5$, is not simple, where $p$ is prime. [Hint: Consider three cases : $p=7$, $p=11$ and $p>11$.]
My attempt :
Let $G$ be a group. $|G| =12p$. Note that $12p=2^2\cdot3\cdot p$. We then have $\text{Syl}_{p}(G)$, $\text{Syl}_{3}(G)$, and $\text{Syl}_{2}(G)$
Now, we have :
$\text{Syl}_{p}(G)$
$n_{p} \big| 12$ and $n_{p} \equiv 1\pmod p$
$\text{Syl}_{3}(G)$
$n_{3} \big| 4p$ and $n_{3} \equiv 1\pmod 2$
$\text{Syl}_{2}(G)$
$n_{2} \big| 3p$ and $n_{2} \equiv 1\pmod 3$
I do not know what to do next. I am not sure if I am on the right track. Thank you in advance.
Consider $n_p$, the number of $p$-Sylow subgroups. $n_p$ satisfies $n_p\mid 12$ and $n_p\equiv 1\pmod p$. If $p=7$ or $p>11$, the only possibility of $n_p$ is 1, so $G$ has a non-trivial normal subgroup.
If $p=11$, $n_p$ is either 1 or 12, $n_2=1$ or $n_2\ge 3$ and $n_3$ is either 1 or greater than or equal to 4. We are going to rule out the case $n_{11}=12$ $n_2\ge3$ and $n_3\ge 4$. If $n_{11}=12$ then $G$ has 12 11-Sylow subgroups $P_1$, $\cdots$, $P_{12}$. We can observe that if $P_i\cap P_j$ contains an non-trivial element then they are coincide because every element of $P_i$ except the identity generates whole $P_i$ so $P_i$ and $P_j$ shares same generator. Therefore $P_i\cap P_j = \{1\}$ if $P_i\neq P_j$. We can argue similarly for Sylow 2- and 3-subgroups so by counting we have $$|G|\ge 1+12\cdot 10+4\cdot2+3\cdot3$$ (1 is for the identity and $12\cdot 10$ is for non-trivial elements of Sylow 11-subgroups $P_i$.) but it is impossible so one of $n_{11}$, $n_3$ and $n_2$ is 1.