#### Prove that if |G| = 2376 then G is not simple .

SOLUTION

$$|G| = 2376 = 2^3 \times 3^3 \times 11$$

If $$n_{11} = 12 \\ \\ Let \ , H \in Syl_{11}(G) \ then \ consider \ \ N_G (H) ; [ G : N_G(H) ] \\ n_{11} = 12 \\ \Rightarrow | N_G(H) | = \frac {2376}{12} = 198 \\ \\ \Rightarrow [ N_G(H) : C_G9H) ] \mid |Aut H | =10 \\ \Rightarrow |C_G(H) | =99 \ or \ 198$$

So , $$9 \mid |C_G(H) |$$ in either case .

So , $$C_G(H)$$ has a Sylow 3- subgroups P (say) $$\Rightarrow$$ P commutes every element of H [ $$as P \leq C_G(H)$$ ]

Now , consider ,

$$H \leq C_G(P) \leq N_G(P) \leq G \Rightarrow |H| = 11 \mid |N_G(P) |$$ .

We again have that this P Sylow -3 -subgroup of $$C_G(H)$$ is a subgroup of a Sylow -3 -subgrou Q (say) of G .

Now , $$[ Q : P ] = 3 \Rightarrow P \leq Q \\ \Rightarrow Q \leq N_G(P) \\ \Rightarrow |Q| = 27 | N_G(P)$$ .

So , we have a subgroup R (say) $$N_G(P)$$

which is divisible by 27 $$\Rightarrow divisible \ by \ lcm(11 ,27) = 297$$ .

Now $$[G : R ] \leq 8 Now , observe that |G| | 8 . \( \Rightarrow |G| | k! \ \ \forall \ k= 1(1) 8 \\ \Rightarrow G \ can’t \ be \ simple$$