Showing that G is not a simple group. I am trying to show that $G$ is not a simple group if 
(i) the order of $G$ is $8000$ or
(ii) the order of $G$ is $2376$.
For Case (i), $8000=2^6\times5^3$. So $n_5=1,16$. If $n_5=1$ then we are done. If $n_5=16$...
For Case (ii), $2376=2^3\times3^3\times11$. So $n_3=1,4$ or $22$. If $n_3=1$ then we are done. If $n_3=4$ or $n_3=22$... 
How to proceed from here. Any help would be greatly appreciated. Thank you.
 A: I'll leave you as an exercise to show that if $|G|=p^rm$ and $n_p\not\equiv 1$(mod $p^2$) then there are two $p$-Sylow subgroups with intersection of size $p^{r-1}$. So let's assume it's correct and do the first part. We assume $G$ is a group of order $8000$ which is simple. Then $n_5=16$. Clearly $16\not\equiv 1$(mod $25$) so there are two 5-Sylow subgroups $P,Q$ such that $|P\cap Q|=5^2=25$. Again, I'll leave you to check that if we have a group of size $p^k$ (where $p$ is a prime) then any subgroup of size $p^{k-1}$ is normal in it. Hence in our case $P\cap Q$ is normal in both $P$ and $Q$. From here we get that for each $g\in P$ we have $g(P\cap Q)g^{-1}=P\cap Q$. Same about $g\in Q$. So that means $P,Q\leq N_G(P\cap Q)$. And that implies $PQ\subset N_G(P\cap Q)$. Hence:
$|N_G(P\cap Q)|\geq |PQ|=\frac{|P||Q|}{|P\cap Q|}=\frac{5^3\times5^3}{5^2}=5^4$
Alright, so what do we know about $|N_G(P\cap Q)|$? It divides $8000$, it is divisible by $5^3$ (because $P$ and $Q$ its subgroups) and it's at least $5^4$. So $|N_G(P\cap Q)|\in\{2^3\times5^3,2^4\times5^3,2^5\times5^3,2^6\times5^3\}$. Now we'll contradict everything. If $|N_G(P\cap Q)|=2^6\times5^3$ then $N_G(P\cap Q)=G$, so $P\cap Q$ is normal in $G$ which contradicts our assumption that $G$ is simple. Now to make it easier let $H=N_G(P\cap Q)$.
Assume $|H|=2^3\times 5^3=1000$. Then $H$ has index $8$ in $G$. We can define an action of $G$ on $G/H$ (the left cosets) by $g.xH=gxH$. An action gives us a homomorphism $\varphi:G\to S_{G/H}$ by $\varphi(g)(xH)=gxH$. As $G$ is simple the homomorphism is either trivial or injective. Easy to see it's not trivial. But it also can't be injective because if it is then $G$ is isomorphic to a subgroup of $S_{G/H}$ and hence by Lagrange's theorem we get that $8000$ divides $8!$ which is a contradiction. 
Now assume $|H|=2^4\times 5^3$. Then the index of $H$ in $G$ is $4$. Define the same action of $G$ on $G/H$ as before. Again we'll get a homomorphism $\varphi:G\to S_{G/H}$. $G$ is simple so it must be trivial or injective. Easy to see it's not trivial, and of course it can't be injective because $G$ has $8000$ elements while $S_{G/H}$ has $24$. A contradiction once again.
Finally assume $|H|=2^5\times 5^3$. Then the index of $H$ in $G$ is $2$. A subgroup of index $2$ must be normal so $H$ is normal in $G$. Again, it contradicts the assumption that $G$ is simple. 
So now try to prove the statements that I left as exercise, and try the second part. 
