If $G$ is a finite group having a $p$-complement for every prime $p$, then $G$ is solvable.
Below shows part of the proof:
Assume that $G$ is simple. Let $|G|=p_1^{e_1}\dots p_n^{e_n}$, where the $p_i$ are distinct primes and $e_i>0$ for all $i$. For each $i$, let $H_i$be a Hall $p_i'$-subgroup of $G$, so that $[G:H_i]=p_i^{e_i}$ and thus $|H_i|=\prod_{j\neq i}p_j^{e_j}$. If $D=H_3\cap\dots \cap H_n$, then $[G:D]=\prod_{i=3}^np_i^{e_i},$ and so $|D|=p_1^{e_1}p_2^{e_2}$. Now $D$ is a solvable group, by Burnside's Theorem. If $N$ is a minimal normal subgroup of $D$, then $N$ is elementary abelian; for notation, assume that $N$ is a $p_1$-group. $[G:D\cap H_2]=\prod_{i=2}^np_2^{e_i}$, so that $|D\cap H_2|=p_1^{e_1}$ and $D\cap H_2$ is a Sylow $p_1$-subgroup of $D$. $N\leq D\cap H_2$ and so $N\leq H_2$. But, as above, $|D\cap H_1|=p_2^{e_2}$, and comparison of orders gives $G=H_2(D\cap H_1)$.
I can't understand the highlighted statement. By the product formula, I am able to prove that $|G|=|H_2(D\cap H_1)|$, but I have no way to prove that $H_2(D\cap H_1)\leq G$, also I can't get the information that $H_2$ or $D\cap H_1$ is normal subgroup of $G$. Is this key hidden in some previous statement?
Now take a subset $S$ of $G$ with $|S|=|G|$. Do you really need $S\leq G$ in order to get $S=G$?