Prove that if |G| = 8000 then G is not simple .

SOLUTION

If \( |G| = 2^3 \times 10^3 = 2^6 \times 5^3 \\ consider \ , \ n_5 = (5k+1) | 2^6 \\ n_5 = 1 , 16 \\ n_2 = (2k +1) | |G| \\ \Rightarrow n_2 = 5 \ , 25 ,\ 125 \) .

Let , H and K are two Sylow – 5- subgroups

\( |H \cap K | | |H| = 5^3 \\ \Rightarrow |H \cap K | = 1 \, 5 , \ 25 \)

If \( |H \cap K| = 1 \\ \Rightarrow |HK| = \frac {|H||K|}{|H \cap K |} = (125)^2 > |G| \)

\( \\ \Rightarrow \Leftarrow \\ \)

If \( H \cap K = 5 \ \\ \ If \ n_5 = 16 \ and \ n_5 = 16 \nequiv 1 \ (mod 11^2) \) .

So there are Sylow- p – subgroups H and K are \( H \cap K \) is of index p in both H and K . Hence normal in each .

So \( N_H(H \cap K) = H ; \ N_K(H \cap K ) = K \\ \Rightarrow | N_G(H \cap K )| > 125 [as \ H \cap K \subset N_G(H \cap K) ] \\ \Rightarrow 125| \ |N_G(H \cap K) \ and \ |N_G (H \cap K )| > 125 \) .

By the …………………………….. \( |N_G(H \cap K)| = 125 \times k \)

althogh this will not be required .

\(N_G(H \cap K ) = 125 \times 8 \rightarrow for \ this \ case \)

\( [ G: N_G(H \cap K ) = 8 \ but \ |G| \not| 8! \ \ \Rightarrow G \ is \ not \ simple \\ N_G(H \cap K) = 125 \times 2^4 \rightarrow for \ this \ case ] \\ [ G:N_G(H \cap K) = 4 \ but \ |G| \not| 4! \Rightarrow G \ is \ not \ simple \\ N_G(H \cap K) = 125 \times 2^5 \rightarrow for \ this \ case ] \\ [G:N_G(H \cap K) =2 \Rightarrow N_G(H \cap K) … G \Rightarrow G \ is \ not \ simple \\ |N_G(H \cap K) |= 125 \times 2^6 \\ =|G| \rightarrow for \ this \ case \\ \\ \\ \\ \\ H \cap K …G \\ \Rightarrow G \ is \ not \ simple \)