# 4 questions from Sylow’s theorem: Qn 3

#### 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$

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