Problem: Prove there are no simple groups of order $1,000,000$.
So, I have used Sylow's Theorem to get that the number of Sylow $5$-subgroups is either $1$ (and we're done) or $16$.
I am assuming we have $16$, and let $H$ and $K$ be two of these Sylow $5$-subgroups. I'm trying to show $H \cap K$ is normal in $G$.
So, $|H\cap K| = |H||K|/|HK| > |H||K|/|G| \approx 244.14$
So since $H\cap K<H$, $|H\cap K|$ divides $|H|$. Thus, $|H \cap K| = 625$ or $3125$.
IF $|H\cap K| = 3125$, then $[H \cap K:H] = 5$ which is the smallest prime divisor of $15625$. So $H \cap K$ is normal in $H$ and normal in $K$. This forces $HK$ to be contained in the normalizer of $H \cap K$. Since the normalizer of $H \cap K$ is a subgroup of $G$, then its order must divide $G$. The size of $HK$ is $3125^2$, which is contained in the normalizer. So $3125^2$ must divide $1,000,000$. Contradiction.
So the normalizer of $H \cap K$ is all of $G$, and $H \cap K$ is normal.
I am completely stuck on what to do if $H \cap K$ has order $625$. Any help would be much appreciated.
With $16$ Sylow $5$-subgroups, your group $G$ embeds into $S_{16}$ (indeed into $A_{16}$). But $10^6\nmid 16!$.