Analysis of Groups of Order 2205 (6.2.6)

Dummit and Foote Abstract Algebra, section 6.2, exercise 6 (excerpt):

Prove there are no simple groups of order $2205=3^2 \cdot 5 \cdot 7^2$.

Proof: Assume $G$ to be a contradiction. By Sylow analysis, we have: $$n_3∈\{7,49\}~~~~~n_5∈\{21,441\}~~~~~n_7=15$$ Index considerations disallow $n_3=7$, though we need only look at $n_7=15$. Since $15 \not ≡ 1 \mod{7^2}$, set $P_0=P_1∩P_2$ for $P_1,P_2∈Syl_7(G)$ such that $|P_0|=7$. We have $P_0 \trianglelefteq P_1, P_2$ so that $7^2 \mid |N_G(P_0)| ≠ 7^2$. Due to index considerations once more, we have $|N_G(P_0)|=3 \cdot 7^2$. But in this case $n_7(N_G(P_0))=1$, so that there is a unique Sylow 7-subgroup of $N_G(P_0)$, despite $P_1$ and $P_2$ being distinct groups of order $7^2$ contained therein.$~\square$