I cann't prove the following statement (Dummit, Foote, Abstract algebra 4.5.17)
Prove that if $\left|G\right|=105$ then $G$ has a normal Sylow $5-$subgroup and a normal Sylow a $7-$subgroup.
It simply to prove that $G$ has a normal Sylow $5-$subgroup or a normal Sylow a $7-$subgroup:
The number of Sylow $5-$subgroup is of the form $n_5=(1+5k)$ Assume that $G$ hasn't a normal then $k\neq 0.$ Note that $n_5 | 21.$ Hence $k=4$ and $n_5=21.$ Then the number of elements g with $\left|g\right|=5, (g \in G),$ is $21\cdot4=84.$
The number of Sylow $7-$supgroup is of the form $n_7=(1+7k).$ Assume that $G$ hasn't a normal then $k\neq 0.$ Note that $n_7 | 15.$ Hence $k=2$ and $n_7=15.$ Then the number of elements $g$ such that $\left|g\right|=7, (g \in G),$ is $15\cdot6=90.$
Since $90+84>105,$ then $G$ has a normal Sylow $5-$subgroup or a normal Sylow a $7-$subgroup.
But why does $G$ have a normal Sylow $5-$subgroup and a normal Sylow a $7-$subgroup?
Best Answer
It's subgroup not "supgroup".
Suppose that $G$ has a normal Sylow $5$-subgroup $N$. Then $|G/N|=21$ and $G/N$ has a normal Sylow $7$-subgroup $M/N$. Then $|M|=35$, so $M$ has a normal Sylow $7$-subgroup $P$. Since $M \le N_G(P)$ with $|G:M|=3$, $|G:N_G(P)|=1$ or $3$. But we cannot have $n_7=3$, so $N_G(P)=G$ and $G$ has a normal Sylow $7$-subgroup.
Alternatively, $P$ is a characteristic subgroup opf $M$ which is normal in $G$, so $P$ is normal in $G$.
The proof in the other case is similar.