Simple group of order 10000 
Prove that no group of order 10,000 is simple.

Here is my attempt.

For contradiction, suppose $G$ of order $10000$ is simple. Notice that
$$10,000 = 5^4 \cdot 16.$$
By the Sylow theorems, $n_5 = 1 + 5k$ for $k \geq 0$ and $n_5 \mid 16$. Going through the cases:
\begin{align*}
k = 0 \implies n_5 = 1 & & \text{divides $16$} \\
k = 1 \implies n_5 = 6 & & \text{doesn't divide $16$} \\
k = 2 \implies n_5 = 11 & & \text{doesn't divide $16$} \\
k = 3 \implies n_5 = 16 & & \text{divides $16$} \\
k = 4 \implies n_5 = 20 & & \text{doesn't divide $16$}
\end{align*}
So $n_5 = 1$ or $n_5 = 16$, but since we're taking $G$ to be simple, we have $n_5 \neq 1$, so $n_5 = 16$.

At this point I know that I need to construct an isomorphism between $G$ and some subgroup of $S_{16}$, using conjugation, and then use Lagrange's theorem. but I cannot figure out how to do this.
 A: If using the action on $G$ by conjugation is slightly difficult for you to grasp, here might be another lengthier method to go about solving this question. (I hope its right)!
Let $n_5=16$. Let $P$ and $Q$ be two sylow-5 subgroups of order $625$ each. Now we know
$$|P \cap Q|=\frac{|P|.|Q|}{|PQ|} \geq \frac{|P|.|Q|}{|G|} \sim 39.06$$
But we know that $|P \cap Q|$ divides 625 as $P \cap Q \subset P$ and so $|P \cap Q|=125$.
This implies $|PQ|=3125$
Now as $[P:P \cap Q]=[Q: P \cap Q]=5$, which is the smallest prime divisor of 625, $P \cap Q \triangleleft P$ and $P \cap Q \triangleleft Q$.
Let $N=N_G(P \cap Q)$. Then, as $P \cap Q$ is normal in both P and Q, we have $P \subset N$ and $Q \subset N$. Now,
$p(P \cap Q)p^{-1}=P \cap Q \ \ \ \forall p \in P$ (as $P \cap Q \triangleleft P$)
and
$q(P \cap Q)q^{-1}=P \cap Q \ \ \ \forall q \in Q$ (as $P \cap Q \triangleleft Q$).
Therefore, for any $pq \in PQ$, we have
$$pq(P \cap Q)(pq)^{-1}=pq(P \cap Q)q^{-1}p^{-1}=p(P \cap Q)p^{-1}=P \cap Q$$
and so we have $PQ \subset N$ and $|PQ| \leq |N|$.
So $|N| \geq 3125$, and it must also be a divisor of $10,000$ and divisible by $625$ (as $P \subset N$ and $Q \subset N$). The only possibilities then are $5000$ and $10000$.
If $|N|=5000$, then its index in $G$ must be 2 and hence it is normal. If $|N|=10000$, then $P \cap Q$ is normal in $G$. Therefore $G$ cannot be simple.
