Could someone please check my solution to this question.
Let G be a group with order 3675. Assume, for contradiction, that G is simple.
$3675=3*5^2*7^2.$
By Sylow 3 we know that the possible number of Sylow p-subgroup in G i.e. $|Syl_p(G)|$ are:
$$ |Syl_3(G)|=7|25|49|1225\\|Syl_7(G)|=15 \\|Syl_5(G)|=21.$$
Because G is simple there is an injection $\varphi:G\to S_{G/H}\cong S_n$. Let $P\in Syl_7(G)$ where $|P|=7^2 $. Then $|P|\Big\vert|G|$ and so $|P|\Big\vert|S_n|$ which implies $n\geq 14$. This means that G has no subgroups of index $\leq 14$ so $|Syl_3(G)|\not=7$.
We consider the normalizers of the larger p-Sylow for the larger primes. By the Orbit-Stabilizer theorem we find that if again $P\in Syl_7(G)$ and $Q\in Syl_5(G)$: $$|N_G(P)|=5*7^2 \ \ \ |N_G(Q)|=5^2*7.$$
These normalizers are subgroups of G and most definitely finite, let $A=|N_G(P)|$ and consider its Sylow p-subgroups. By applying Sylow 3 again we find:
$$ |Syl_7(A)|=1 \\ |Syl_5(A)|=1.$$
Take $R\in Syl_5(A)$, since there is only one such $R$ it follows that $R⊲A$. And since $R\subseteq A\implies A\subseteq N_G(R)$. But $N_G(R)$ we know from before is $|N_G(Q)|=5^2*7$. And since $A\subseteq |N_G(Q)|$, $A$ must by Lagrange also divide $|N_G(G)|$. This leads us to the statement: $$|A|\Big\vert|N_G(G) \leftrightarrow 5*7^2\Big\vert5^2*7 $$
This is a contradiction. Both $|Syl_5(G)|=15$ and $|Syl_7(G)|=21$ cannot be true at the same time which implies that G is not simple.
Suppose that $G$ is a simple group of order $3675 = 3 \cdot 5^2 \cdot 7^2$. Then $n_7(G) = 15$. First suppose that $P$ and $Q$ are two Sylow $7$-subgroups with $|P \cap Q| = 7$. Since $P \cap Q \lhd P$ and $P \cap Q \lhd Q$, it follows that $P,Q \subseteq N_G(P \cap Q)$. Thus $7^2$ divides $|N_G(P \cap Q)|$ and $N_G(P \cap Q)$ contains at least two Sylow $7$-subgroups. Then $N_G(P \cap Q)$ must contain $15$ Sylow subgroups. Thus $3 \cdot 5 \cdot 7^2 \mid |N_G(P \cap Q)|$. Then $[G:N_G(P \cap Q)] \le 5$, which is impossible because $|G| > 5!$.
Therefore, $|P \cap Q| = 1$ for every pair of distinct Sylow $7$-subgroups. It follows that $n_7(G) \equiv 1 \bmod 7^2$, a contradiction because $n_7(G) = 15$.
This last congruence follows from a general result. Here is the argument for this case. Let $P$ be a Sylow $7$-subgroup and let $P$ act by conjugation on the set of Sylow $7$-subgroups. Then one orbit has size 1 (the orbit containing $P$) and the remaining orbits have size $7^2$. To see this, let $Q$ be a Sylow $7$-subgroup distinct from $P$. Then the size of the orbit containing $Q$ is $\frac{|P|}{|N_P(Q)|}$. We will show that $P \cap Q = N_P(Q)$. Clearly $P \cap Q \subseteq N_P(Q)$. Let $x \in N_P(Q)$. Then $x$ has order a power of $7$ and $x Q x^{-1} = Q$, which implies that $\langle Q, x \rangle$ is a $7$-group. Since $Q$ is a $7$-group of maximal size, it follows that $x \in Q$ and thus $x \in P \cap Q$. Putting this together gives $\frac{|P|}{|N_P(Q)|} = \frac{|P|}{|P \cap Q|} = \frac{49}{1} = 49$.