A "full" solution set means:

(specifically Group Actions, the Sylow Theorems, and the Jordan-Hölder Theorem), represents the point where the subject moves from basic definitions to profound structural analysis. 1. The Pedagogical Weight of Chapter 4

: Groups Acting on Themselves by Left Multiplication (Cayley’s Theorem). Section 4.3

\beginproof $n_7 \equiv 1 \pmod7$ and $n_7 \mid 8$, so $n_7=1$ or $8$. If $n_7=1$, the Sylow $7$-subgroup is normal. If $n_7=8$, then $8(7-1)=48$ elements of order $7$. The remaining $56-48=8$ elements form the Sylow $2$-subgroups; each Sylow $2$-subgroup has order $8$. But then $n_2 \mid 7$ and $n_2\equiv 1 \pmod2$, so $n_2=1$ or $7$. $n_2=1$ gives a normal subgroup. $n_2=7$ gives $7$ subgroups of order $8$, each containing identity, total elements $7\cdot 7 +1$? Let's check carefully: the intersection of distinct Sylow $2$-subgroups can be large; but a standard argument: if $n_7=8$, then the normalizer of a Sylow $7$ has index $8$, so $|N_G(P_7)|=7$. But $P_7$ is cyclic of order $7$, so $N_G(P_7)$ contains $P_7$ and possibly an element of order $2$ (since $56/7=8$, the normalizer size is $7$ or $56$; if $n_7=8$, then $|N_G(P_7)|=7$, so no element of order $2$ normalizes $P_7$, contradiction to counting). Thus $n_7$ cannot be $8$. Hence $n_7=1$, so $G$ not simple. \endproof

\beginproblem[Exercise 4.3.5] Show that if $G$ is a group of order $p^2$ ($p$ prime), then $G$ is abelian. \endproblem