This is a problem from the exam I took this morning
"Prove there is no simple group of order $2^73^2.$"
This is what I do, suppose there is such a group $G$. Then by Sylow's theorem, $n_2=3,9$. If $n_2=3$, then we have a map $G\to S_3$. The kernel cannot be trivial since $G$ is too big, so this case is done. If $n_2=9$, then there is a map $G\to S_9$ and since $G$ is not the zero map and $G$ is simple, then this map is injective. This is where I'm stuck.
This is a Qual exam, so I cannot use big theorems that trivialize the problem. So Burnside pq theorem cannot be used. How do I solve this prob?
Continuing from where you left.
Let $H$ be the subgroup of $S_9$ isomorphic to $G$. (That is, the image of the injective homomorphism that you got.)
Note that $|H| > 2$ and thus, $H \cap A_9$ is nontrivial. (Can prove this easily via contradiction by assuming that every non-identity permutation of $H$ is odd and considering the product of two elements which aren't inverses of each other.)
Now, set $K = H \cap A_9$. As $A_9 \unlhd S_9$, we get that $K \unlhd H$.
By our assumption, $G$ is simple and hence, so is $H$. As we have eliminated the option that $K = (1)$, this forces $K = H$.
Thus, $H \cap A_9 = H$ and hence, $H \le A_9.$
However, observe that $|A_9| = \frac{9!}{2} = 2^6\times(\text{odd}).$ Thus, $|H| \not\mid |A_9|,$ contradicting Lagrange's Theorem.