D+F 4.5.13

Prove that a group of order 56 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.

$\textit{Proof}$: First note that $56=8\cdot 7 = 2^3 \cdot 7$. By part (iii) of Sylow's Theorem, the number of Sylow $p$-subgroups of $G$ is $$n_p \equiv 1(modp)$$ Additionally, $$n_p \mid m $$ The prime components of 56 are 2 and 7. So, $$n_2 \equiv 1(mod2)$$ and $$n_2 \mid 7$$ If we generate a list of possible values for $n_2$, we get $$n_2 = 1,7$$ If we generate a list of possible values for $n_7$, we get $$n_7 = 1,8$$ We will proceed by a counting argument, then using corollary 20 on page 142. Consider the case when $n_2=7$ and $n_7=8$. In other words, there are 7 subgroups of order 2 and 8 subgroups of order 7. Groups of order 7 and 2 are isomorphic to the cyclic groups $\mathbb{Z}_7$ and $\mathbb{Z}_2$, respectively. $\mathbb{Z}_7$ has 6 non-identity elements (of order 7) and $\mathbb{Z}_2$ has 1 non-identity element. In total, there are $8 \cdot 6 = 48$ elements of order 7 and $7 \cdot 1 = 7$ elements of order 2. Counting up, this gives us $48+7+1=56$ elements in $G$. However, when multiplying elements together in $G$, if $|a|=7$ and $|b|=2$, then $|ab|=lcm(7,2)=14$. So, there must be more elements in $G$ with order 14. This is a contradiction since $|G|=56$. Therefore, it must be the case that either $n_2=1$ or $n_7=1$. By the corollary on p. 142, this is equivalent to one of these Sylow $p$-subgroups being normal.
So, by a counting argument, we have forced that there must exist a normal Sylow $p$-subgroup of $G$ when $|G|=56$. $\blacksquare$