Suppose that $G$ is a group with order $2000$. Prove that the group is not simple. How about when the group order is $4000$?
Here is my work so far, for the order $2000$ case:
Note that $2000 = 2^45^3$.
Suppose that $G$ is simple. Then all of its Sylow groups are not normal in $G$. Hence by Sylow theorem $n_5 := |Syl_5(G)| = [G:N_G(P_5)] = 1 + 5k \space| 2^4 = 16$ for some $k$ a natural number not zero (here, $P_5$ is a Sylow-$5$ group). It follows that $n_5 = 16$, so that by Lagrange's Theorem $|N_G(P_5)| = 125.$ Let $P$ and $Q$ be distinct Sylow-$5$ groups, and let $I :=P \cap Q$. Notice that the set $PQ:=\{pq \space|\space p \in P, q\in Q\}$ is a subset of $G$ and hence must have an order at most $2000$. We know that $|PQ| = \frac{|P||Q|}{|P \cap Q| }$. Because $I$ is a subgroup of $P$ and $P$ has an order of $125$, by Lagrange's theorem we have that $|I| = 1, 5, 25$ (notice that it cannot be 125 since $P$ and $Q$ are distinct). If $|I| = 1$ or $5$, then $|PQ| \geq 5^6$ or $5^5$, respectively, both of which are greater than $2000$, a contradiction. Hence it must be that $|I| = 25$.
But I wasn't sure how to go about from here. I've already considered $n_2 := |Syl_2(G)|$ as well, but that didn't seem to work either. Maybe I should consider other directions. My professor was hinting that I might have to use the following theorem:
Suppose $G$ is a finite group and let $H$ be a $p$-subgroup of $G$. If $p \space | \space [G:H]$, then $p \space | \space [G: N_G(H)]$.
But I wasn't sure how to use this.
I am thinking that a similar method for solving $2000$ might work for $4000$ as well, but I haven't got the case of $4000$ either.
I also want to prove this without using Burnside's theorem.
You got as far as $|I|=25$. Note that $I$ has index $5$ in both $P$ and $Q$, so it is normal in them both. So $N_G(I)$ contains more than one Sylow $5$-subgroup, and hence it contains at least $16$. So $|N_G(I)|$ is divisible by $125$ and by $16$.
So, for $|G|=2000$, $N_G(I) = G$, and $G$ is not simple. For $|G|=4000$, $N_G(I)$ could have index $2$ in $G$, but then again $G$ is not simple.