If $G$ is a group of order $|G|=10^6$, then $G$ cannot be simple.
I hame hoping to get feedback on my proof attempt; I am using:
Lemma: Let $|G|=p^am$, where $a>1$ and $m>1$. Assume $G$ is simple, then the following are satisfied:
- $n|m$
- $n \equiv 1 \pmod p$
- $|G|$ divides $n!$ where $n=n_p(G)$
Proof: Assume $G$ is simple with $|G|=10^6=2^6 \cdot 5^6$. We can apply the Lemma from above with $p=5$. Let $n=n_5(G)$. Then, $n\mid 2^6=64 \implies n \in \{2^1,2^2,2^3,2^4,2^5,2^6\}$. Since $n \equiv 1(\mod 5)$ and $10^6\mid n!$, then $2^4 \equiv 1(\mod 5)$ (no other powers of 2 fit this way). Since we assumed that $G$ is not simple, then $n \neq 1$. Thus, $n=2^4=16$. To see if $10^6\mid 16!$, we can look at $5^6\mid 16!$. $16!=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot ... \cdot 10 \cdot ... \cdot 15 \cdot 16$. Clearly, there are three $5$'s contained in $16!$.
$\therefore$ The largest power of $5$ to divide $16!$ is 3.
$\therefore$ $G$ is not simple.