(a) Prove that a Sylow $7$-subgroup of $G$ is normal
(b) Prove that $G$ is Solvable
Can anyone please tell me if I am correct?
(a) For the sake of contradiction suppose $G$ dose not have a normal Sylow $7$-subgroup.
We first show $G$ has a normal Sylow $5$-subgroup. Then $G$ must have $15$ Sylow $7$-subgroups. So $G$ has $15(7-1) = 90$ elements of order $7$. If $G$ dose not have a normal Sylow $5$-subgroup then $G$ has $21$ Sylow $5$-subgroups so $G$ has $21(5-1) = 84$ elements of order $5$. But $90 + 84 = 174 > 105$. Therefore $G$ has a normal Sylow $5$-subgroup.
Let $N$ be the unique Sylow $5$-subroup, and let $P$ be a Sylow $7$-subgroup. Since $N$ is normal $NP$ is a subgroup of $G$. Since $N \cap P = 1$ we have $|NP| = |N||P| = 35$. So by Lagrange $|G : NP| = 3$ since $3$ is the smallest prime dividing $|G|$ we have that $NP$ is normal. So the Fratini Argument $G = N_G(P)N$
Finally since $NP$ is abelian $NP$ normalizes $P$. So $NP \leq N_G(P)$ Bur since $3$ divides $|G|$ and $3$ dose not divide $N$ we have $3$ divides $N_G(P)$ so $105$ divides $N_G(P)$ thus $G = N_G(P)$.
(b) Continuing with the notation above $NP$ is a normal subgroup of $G$ and $G/NP$ has order $3$ so is clearly abelian. Since $NP$ is a abelian, the trivial subgroup $1$ is a normal subgroup of $NP$ and $NP/1$ is abelian. Hence $1 < NP < G$ is our disired chain.
Also if anyone has any nice rules for proving that groups of a certain order are solvable that would be appreciated. I herd groups with order divisible by at most $2$ distinct primes must be solvable.