How to prove that there is no simple group of order 10 000 000?
Here is what I have so far: Let $G$ be such a group, we have $|G|=10 000 000=2^7*5^7$.
To prove that $G$ is not simple a solution could be to prove that there is $p \in \{2,5\}$ such that the number of p-Sylows is one.
We have $n_2 \equiv 1 (2)$ and $n_2 | 5^7=78125$ but then counting the number of possibilities for $n_2$ is rather long.
Similarly, $n_5 \equiv 1 (5)$ and $n_5 | 128$. It is easier to count the possible values of $n_5$: $n_5$ is either 1, either 16. As $G$ is simple we can assume it is 16. Then, there are $16*(5^7-1)=1 249 984$ elements of order 5, 25, ..., 78125.
Then the smallest possible value for $n_2$ that is greater than 1 (for $G$ to be simple) is 5 as $5 | 5^7$ and $5 \equiv 1 (2)$ and there are $5*(2^7-1)=635$. So actually $n_2=5$ could work, I must have made a mistake somewhere.
Assume that $G$ is simple and that $n_5 = 16$. By Sylow's third theorem, $n_5 = 16 = [ G \ : \ N_G(P) ]$, where $P \in Syl_5(G)$.
Cayley's theorem states that for any $H < G$ there is a homomorphism $f : G \rightarrow S_n$, where $n = [ G \ : \ H ]$ such that $Ker(f) \leq \cap_{g \in G} H^g$.
Apply Cayley's theorem to $H = N_G(P)$. Since $G$ is simple, it follows that $Ker(f) = \{1\}$ and thus $|G| \ \big| \ |S_{16}| = 16!$, which gives a contradiction.