The Millionaire (6.2.28)

Dummit and Foote Abstract Algebra, section 6.2, exercises 28:

Let $G$ be simple and of order $3^3*7*13*409=1,004,913$. Calculate the number of Sylow $p$-subgroups for each prime dividing $|G|$.

Proof: Preliminary Sylow analysis shows:$$n_3∈\{7,13,91,409,2863,5317,37219\}~~~~~n_7∈\{351,47853\}$$$$~~~~~n_{13}∈\{27,25767\}~~~~~n_{409}∈\{819\}$$Removing the numbers that violate the index restrictions, we have all of them solved except for $n_3$. Assume $n_3=409$, and we have $|N_G(P_3)|=3^3*7*13$, so for some $P_7$ we have $P_3P_7$ is a subgroup of $N_G(P_3)$, so that naturally $P_7 \trianglelefteq P_3P_7$, meaning $|P_3P_7|=189 \mid |N_G(P_7)|=21$, a contradiction. Assuming $n_3=5317$, we end up with the same contradiction as then $|N_G(P_3)|=3^3*7$.

A brief digression: By counting the elements of order 7, 13, and 409, we obtain $930,474$ elements. If there is an element $x$ of order 21, then $P_7 \trianglelefteq \langle~x~\rangle$ so that $\langle~x~\rangle = N_G(P_7)$ by order, so that by 6.2.16's first lemma there are $12*47853=574,236$ elements of order 21, overloading $G$.

Now assume $n_3=37219$, and since $37129 \not ≡ 1 \mod{3^2}$, we have an order-$3^2$ intersection of two Sylow 3-subgroups $P_3∩Q_3$, whose normalizer is of order $3^3*7$,$3^3*13$, or $3^3*7*13$. If $3^3*7$, then by the above the only order the elements in this normalizer can take are 1, 7, and powers of 3. Sylow shows $n_3=7$ (by 6.2.13) and $n_7=1$, so that there are $3^3*7-6=183$ elements in the Sylow 3-subgroups. Without taking intersections into account, there is a maximum of $7*26+1=183$ elements possible, so that there is no intersection between Sylow 3-subgroups taking place, despite $P_3,Q_3 ≤ N_G(P_3∩Q_3)$ and $P_3∩Q_3 ≠ 1$, or 6.2.13 in general. If $3^3*13$ or $3^3*7*13$, the order of the normalizer forces only one Sylow 3-subgroup, despite once again $P_3,Q_3 ≤ N_G(P_3∩Q_3)$ being distinct Sylow 3-subgroups. Therefore, $n_3=2863$ by default.$~\square$