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September 30, 2019

4 questions from Sylow’s theorem: Qn 1

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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