Exercise 4.5.17 (If $|G| = 105$ then $G$ has a normal Sylow $5$-subgroup and a normal Sylow $7$-subgroup)

Proof. Suppose that $|G|=105=3\cdot 5\cdot 7$.

By Sylow’s Theorem, $n_5\mid 21$ and $n_5\equiv 1(\mathit{𝑚𝑜𝑑}5)$, $n_7\mid 15$ and $n_7\equiv 1(\mathit{𝑚𝑜𝑑}7)$, thus

We show first that $n_5=1$ or $n_7=1$. If not, $n_5=21$ and $n_7=15$. Therefore there are $n_5\cdot 4=21\cdot 4=84$ elements of order $5$ and $n_7\cdot 6=215\cdot 6=90$ elements of order $7$, so $|G|\ge 84+90=174$, in contradiction with $|G|=105$. This proves

Let $P$ be a Sylow $5$-subgroup and let $Q$ be a Sylow $7$-subgroup. Suppose that $n_5=1$ (the case $n_7=1$ is similar). Then $P⊴G$, and $Q\le G$, therefore $\mathit{𝑃𝑄}$ is a subgroup of $G$. Moreover $P\cap Q=\{1\}$, therefore $|\mathit{𝑃𝑄}|=\frac{|P||Q|}{|P\cap Q|}=35$, so $|G:\mathit{𝑃𝑄}|=3$ is the least prime factor of $|G|$. This shows that $\mathit{𝑃𝑄}⊴G$.

Since $|P|=5\nmid 6=|Q|-1$, then $\mathit{𝑃𝑄}$ is cyclic (see the Example, groups of order $\mathit{𝑝𝑞}$, p.143). Therefore $P$ is the unique subgroup of order $5$ in $\mathit{𝑃𝑄}$ and $Q$ is the unique subgroup of order $7$, so $P$ and $Q$ are characteristic subgroups of $\mathit{𝑃𝑄}$, where $\mathit{𝑃𝑄}⊴G$, hence $P⊴G$ and $Q⊴G$.

If $|G|=105$ then $G$ has a normal Sylow $5$-subgroup and a normal Sylow $7$-subgroup. □