Proof that a group of order 8888 is solvable

Question

I need to prove that every group of order $8888$ is solvable by proving that $G^{(3)}=\{e\}$. Can you give some help?

Answer 1

Maybe you should display some efforts of your own. Let me nevertheless give some hints / one way to do this. $8888=2^3 \cdot 11 \cdot 101$. If $P \in Syl_{101}(G)$, prove that $P \lhd G$, since all divisors of $88$ greater than $1$ are smaller than $101$. Look at $G/P$, and show that its Sylow $11$-subgroup is normal (also by inspecting the $\gt 1$ divisors of $8$). If $Q \in Syl_{11}(G)$, this amounts to $PQ \lhd G$. Now use that $PQ$ is cyclic of order $1111$ (and index $8$) and that any group of order $8$ has a trivial second derived subgroup.