Show that there is no simple group of order $3393$
The hint I was given was to look at Sylow $3$-subgroups. I know that $3393 = 3^2 \times 13 \times 29$ and that $G$ must contain a subgroup of order $3^2$, say $H$. But how can I proceed further with this. The proof I managed to come up with was this.
Attempted Proof: Let $G$ be a simple group of order $3393$. Sylow theory tells us that the number of Sylow $29$ subgroups of $G$, $n_{29}$ is either $1$ or $117$ and that the number of Sylow $13$ subgroups of $G$, $n_{13}$ is either $1$ or 260 $261$. Since $G$ is simple we must have $n_{13} = 261$ and $n_{29} = 117$.
Let $\operatorname{Syl}_{29}(G) = \{H_i \ | 1 \leq i \leq 117\}$, since $H_i \cap H_j \leq H_i$ for any $i \neq j$ we must have $H_i \cap H_j = \{1_G\}$, for otherwise we'd end up with $H_i \cap H_j$. This shows that we have at least $117 \times 28 = 3276$ elements of order $29$ in $G$. Applying the same reasoning to $\operatorname{Syl}_{13}(G)$ we see that $G$ must also contain $261 \times 12 = 3132$ elements of order $13$. Thus $|G| \geq 3132 + 3276 > 3393 = |G|$ a contradiction. $\square$
Are there any errors in my proof? If not, how can I use the hint to prove this?
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This is a bit late, but for posterity: there is in fact a sneaky way to do it by looking at the number of Sylow 3-subgroups (source: Dummit & Foote page 203; hopefully I'm exegeting it correctly :) ).
Suppose towards contradiction that $G$ is simple and has order $3393=3^2\cdot 13\cdot 29$. Let $n_3$ denote the number of Sylow 3-subgroups. By Sylow's Theorem $n_3\equiv 1 \mod 3$ and $n_3|13\cdot 29$. If $n_3=1$, we get that that Sylow 3-subgroup is normal and therefore $G$ is not simple. So basically, it must be either 13 or 29, but only 13 is congruent to 1 mod 3, so $n_3=13$. Also by Sylow's Theorem there is some Sylow 3-subgroup whose normalizer $H$ has index 13 in $G$.
(Some details to fill in for justifying the following paragraph, but) we can make a homomorphism $\pi_H: G \rightarrow S_{13}$. (Essentially, the codomain is representing the set containing $13$ cosets of $H$, and each $g\in G$ sends a coset of $H$ to some coset of $H$, so each element $g$ corresponds to a certain permutation of the set of cosets). Now recalling that kernels of homomorphisms are normal subgroups, we consider $\ker(\pi_H)$.
If $\ker(\pi_H)$ were all of $G$, that would mean that $H$ is all of $G$, but it's not because $H$ has index 13. Since $G$ is simple, $\ker(\pi_H)$ must be the trivial group. By the First Isomorphism Theorem, $G/\ker(\pi_H)\cong \img(\pi_H)$. Since $\img(\pi_H)$ is a subgroup of $S_{13}$, we know by Lagrange's Theorem that $|\img(\pi_H)|$ divides $13!$, so $|G|$ divides 13!. However, 29 divides $|G|$ but is not a factor of $13!$. This is inconceivable and outrageous, a contradiction.