Can a group of order 3000 be a simple group? Can a group of order 3000 be a simple group? How about the case of a group of order 1000?
 A: A group of order 1000 has a normal Sylow 5-subgroup, so it cannot be simple.
A group of order 3000 has either a normal Sylow 5-subgroup or 6 of them. If there are 6 Sylow 5-subgroups, then the normalizer of a Sylow 5-subgroup has order 3000/6 = 500. A subgroup of order 500 cannot have a trivial normal core in a group of order 3000: Index of the normal core in the Sylow 5-subgroup must divide (6 - 1)! but 500 does not divide 120 = 5!.
A: Theorem: Let $G$ be a simple group and $H<G$ such that $[G:H]=n$ then $$G\hookrightarrow A_n$$
A: Hints:
(1) Let $\,G\,$ be a group, $\,H\le G\,\;,\;\;X:=$ the set of left cosets of $\,H\,$ in $\,G\,$ . The regular action of $\,G\,$ in $\,X\,$ is given by $\,g(g_1H)\mapsto(gg_1)H\,$ and, as any other action of a group on a set, it determines a group homomorphism $\,\phi:G\to \text{Sym}_X:=\,$ the group of all bijections from the set $\,X\,$ to itself. Note that if $\,[G:H]=n\,$ then $\,\text{Sym}_X\cong S_n\,$ .  The kernel of this homomorphism, say $\,N:=\ker\phi\;$ , is the maximal normal subgroup of $\,G\,$ that is contained in $\,H\,$ .
2) From the above it follows that if a group $\,G\,$ cannot be embedded in some $\,S_n\,$ then $\,|N|>1\,$
3) Thus, a group $\,G\,$ with a subgroup $\,H\le G\,$ s.t. $\,[G:H]=n\,$ and $\,|G|\,\nmid\,n!\,$ cannot be simple
Finally, a group of order $\,3,000=5^3\cdot 3\cdot 3^3\,$ has either one or six Sylow $\,5$-subgroups, yet the second option would there exists a subgroup in $\,G\,$ of index $\,6\,$ , and this means $\,G\,$ cannot be simple.
A: let $G$ be a group of 3000 .Then by SYLOWS theorem $G$ contain a subgroup $H$ of order 1000. as$ 0(G)$ does not devide factorial of $(i(H))$. $H$ contain a non trivial normal subgroup $K$ of $G$ which contain in $H$. so $G$ is not simple. where $i(H)$ means index set of $H$ in $G$  
