Let $G$ be a group of order $1004913 = 3^3 \cdot 7 \cdot 13 \cdot 409$. We suppose that $G$ is simple. We want to obtain a contradiction.
This is the Exercise 29 in Chapter 6.2 of Dummit-Foote. As far as I know, in MSE, there are two QAs: QA1, QA2. And I found an exercise and a solution.
I have determined $n_3, n_7, n_{13}, n_{409}$ uniquely as mentioned in QAs above. We consider the action of $G$ on $Syl_{409}(G)$ by conjugation. Let $G \to S_{819}$ be the permutation representation of this. I see that elements of order $3$ in $N_G(P)$ fix exactly $3$ points in $S_{819}$, where $P \in Syl_{409}(G)$. However, I don’t know how to obtain a contradiction by this.
According to QAs above, elements of order $3$ in $G$ fix exactly $9$ points in $S_{819}$, which follows from Burnside’s lemma or some higher theory. In Dummit-Foote, Burnside’s lemma is stated in the last chapter about character theory. Therefore, I wonder if there is a solution without it, i.e., solution with tools in Chapters 1-6.
P.S. This question does not mean that I hesitate to learn theorems beyond Dummit-Foote Chapters 1-6. I have found that Burnside’s lemma appears in an early chapter in Rotman. I am going to try to learn it later. However, I want to know or guess what the solution that the authors expected is.
Update: After I thought twice again, I solved this exercise as follows. We can show that $P \cap Q = 1$ for any $P, Q \in Syl_{3}(G)$ with $P \neq Q$. By counting elements of $G$, it follows that there is no chance of existence of an element of order other than $1, 3, 3^2, 3^3, 7, 13, 409$. In particular, the number of elements of either order $3$, $3^2$ or $3^3$ is determined as $N = 7 \cdot 409 \cdot 26$. We count the number $M$ of elements of order $3$ in $\bigcup_{P \in Syl_{409}(G)} N_G(P)$. Noting that there are no element of order $3 \cdot 409$, we see that $N_G(P)$ is not cyclic, i.e., each $N_G(P)$ has $409$ Sylow $3$-subgroups. By the fact that elements of order $3$ in $N_G(P)$ fix exactly $3$ points in $S_{819}$, we see that $M = 2 \cdot 409 \cdot 819 / 3$, which gives a contradiction since $M > N$.
Although I solved the exercise, I have not understand other solutions. I will be appreciated if you could tell me answers of the following questions.
- How can we prove that $P$ is elementary Abelian for $P \in Syl_{3}(G)$? I know that it suffices to prove that there is no element of order $9$ or that $P$ is Abelian. For the latter, the N/C theorem tells me that $P$ is elementary Abelian.
- How can we prove that any elements of order $3$ within $P$ fix the same number of points in $S_{819}$? If we can show this, I know that it follows that all the elements of order $3$ in $G$ fix the same number of points in $S_{819}$ since Sylow groups are conjugate.
If I can show item 1., I know how I can use Derek’s hint. If I can show item 2., now I know how I can apply Burnside’s lemma to gain a contradiction.
Update 2: I saw Derek’s second advice and I decided to determine all the five groups of order $p^3$ for odd primes $p$. When I wrote item 1., I knew the isomorphism types of $\mathrm{Aut}(G)$ for the three Abelian groups $G$ of order $p^3$. Now I know $\mathrm{Aut}(G)$ for the two non-Abelian groups, too. Hence, I have managed to prove that $P$ is elementary Abelian for $P \in Syl_{3}(G)$ and to obtain contradiction following Derek's first advice. I would like to thank Derek here.
Actually, I have not seen how to prove item 2. yet. But it's time to move on to the next study.
Let $\left|G\right|=1004913=3^{3}7\cdot 13\cdot 409$ and suppose $G$ is simple. Then the Sylow $3$-subgroup is elementary abelian and the class equation implies there are at most $2$ cycle types for elements of order $3$. On the other hand, using the permutation representation in $S_{819}$, the normalizers for the Sylow $7$, $13$, and $409$ subgroups produce $3$ distinct cycle types for elements of order $3$, a contradiction. It must be that $G$ is not simple. Thus the result.