Simple group of order $3420$ has no element of order $171$ I'm stuck on an old algebra prelim problem which requires me to show that a simple group $G$ of order $3420= 2^2 \cdot 3^2 \cdot 5 \cdot 19$ has no element of order $171= 3^2 \cdot 19$.   
So far, since I know that $G$ has no nontrivial normal subgroup, it is clear that $n_{19}= 20, n_{5}= 6, n_{3} \in \{4, 19,76\}, n_{2} \in \{3,5,9,15,19,45,57,95,171,855\}$ via Sylow's Theorems. The next step would be to consider action of $G$ (by conjugation) on the set of Sylow $p$-subgroups of $G$ for some prime divisor $p$ of $3420$ (given as a hint). But which $p$(s) should I be using to follow this hint? I'd appreciate all the help I can get here, thanks.
 A: I should choose $p=3$ and remember that the Sylow Subgroups are permuted transitively by conjugation. If there are small numbers of subgroups you can get a non-trivial homomorphism to a small symmetric group, where the kernel is normal.
You might want to notice $p=19$ where $20$ is not a multiple of $3$.
Consider the structure of the supposed subgroup of order $171$ in the light of these facts.

Note that the case of $10$ subgroups of order $9$ has not been excluded (see comments) though not mentioned in the question. So this is not a full answer.
A: You can use the following lemma: Assume that $G$ is a simple group, $H\leq G$ and $|G:H|=k>1$. Then $G$ can be considered as a subgroup of $\mathbb S_k$.
For proof, consider the action $\phi:G\longrightarrow\mathrm{Sym}(\{aH\mid a\in G\})\cong \mathbb S_k$ defined with $\phi:g\mapsto \tau_g$, where $\tau_g:aH\mapsto gaH$. Th kernel $\ker(\phi)$ is trivial, since $\ker(\phi)\triangleleft G$, $G$ is simple, and also if $g\notin H$, then obviously $g\notin\ker(\phi)$, so $\ker(\phi)\neq G$. Therefore, $\phi$ is injective, so we can consider $G$ as a subgroup of $\mathbb S_k$.
Assume now that $|G|=2^2\cdot 3^2\cdot 5\cdot 19$ and $G$ is simple. Toward contradiction, assume that $a\in G$ is an element of order $3^2\cdot 19$. Then $|G:\langle a\rangle|= 20$, hence by lemma we can assume that $G\leq\mathbb S_{20}$. So, $a\in\mathbb S_{20}$ is an element of order $3^2\cdot 19$, which is not possible (you can easily see that by considering cycle decomposition of $a$).
