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Prove that if |G| = 616 then G is not simple .

SOLUTION

$|G| = 616 = 2^3 \times 7 \times 11$

Consider the 11 – sylow subgroup of G .

$n_{11} \mid |G| = 616 \ \ n_{11} = number \ of \ sylow_ {11} subgroup \\ \Rightarrow (11k + 1) \mid 56 \ \Rightarrow n \in \{ 1 \ , 56 \}$

k = 0 , 5

$n_{7} \ is \ the \ number \ of \ sylow_ 7 \ subgroup \\\Rightarrow n_{7} = 97k + 1)|88 \ \ k = 0 , 1 , 3 \rightarrow n_{7} \in {1 , 8 , 22}$

$n_{2} =( 2k + 1) | 77 \Rightarrow k = 0 , 3 , 5 , 38 \\ n_{2} \in \{ 1 , 7 , 11 , 77 \}$

Suppose that $n_{7} \neq 1 and \ n_{11} \neq 1$ then for any $H_{1} , H_{2} \in Syl_{11} (G) \\ k_1 , k_2 \in Syl_7 (G)$

we have $|H_1 \cap H_2 | \mid |H_1| , |H_2| \\ \Rightarrow H_1 \cap H_2 = \{ e \} \ any \ k_1 \cap k_2 = \{ e \} \ \ and |H_1 \cap k_1 | \ \mid |H_1| and |k_1 | \\ \Rightarrow |H_1 \cap k_1 | = 1 \Rightarrow H_1 \cap k_1 = \{ e \}$ \)

So we have $56 \times (11 – 1) = 560$ elements of order 11 .

at last $8 \times (7 -1) = 48$ elements of order 7 .

As 616 – 560 – 48 = 8

$\Rightarrow$ G has only one $sylow_2$ subgroup & $\Rightarrow$ That $sylow_1$ subgroup would be normal in G . $\\ \\ \\ \\ \Rightarrow$ G is not simple .
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