Let $G$ be a group in which all maximal non-trivial subgroups are Sylow subgroups. Then $G$ isn't a simple non-abelian group.
I know how to prove this by relying on the theorem that if all maximal subgroups are nilpotent, then the group itself is solvable (in particular, isn't simple non-abelian group).
But I wonder if it's possible to prove this statement in another way?
I will be grateful for any help.
Here's an implementation of the computations that Derek Holt had in mind.
Let $$ |G|=n=p_1^{s_1}\ldots p_r^{s_r}, $$ where $p_i$ are distinct prime numbers and let $P_i$ be a Sylow $p_i$-subgroup. We assume that $N_G(P_i)=P_i$ for all $i$ and $r\geq2$.
Since any pair of Sylow $p_i$-subgroups has a trivial intersection and the number of Sylow $p_i$-subgroups is $$ |G:N_G(P_i)|=|G:P_i|=\frac{n}{p_i^{s_i}}, $$ it follows that the number of $p_i$-elements in $G$ is $$ (p_i^{s_i}-1)\frac{n}{p_i^{s_i}}. $$ Hence $$ n\geq1+\sum_i(p_i^{s_i}-1)\frac{n}{p_i^{s_i}}=1+rn-n\left(\frac{1}{p_1^{s_1}}+\ldots+\frac{1}{p_r^{s_r}}\right)\geq1+rn-n\cdot\frac{r}{2}=1+n\cdot\frac{r}{2}\geq n+1. $$ Contradiction.