# sep05 - unique Sylow 3-subgroup which must be P and we have...

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Math 5125 Monday, September 5 September 5, Ungraded Homework Let A ± G be groups with A abelian and deﬁne θ : G Aut ( A ) by θ g ( a ) = gag - 1 . Prove that ker θ A . If g A , then gag - 1 = a for all a A , because A is abelian. This means that θ ( g ) is the identity, equivalently g ker θ and the result follows. Exercise 4.4.12 on page 138 Let G be a group of order 3825. Prove that if H is a normal subgroup of order 17 in G , then H Z ( G ) . As usual, write 3825 as a product of prime factors: 3825 = 3 2 * 5 2 * 17. Let P be a Sylow 3-subgroup and Q a Sylow 5-subgroup of G . Now PH G because H ± G . By Lagrange P H = 1 and we have PH / H = P / P H = P / 1 = P . It follows that | PH | = 3 2 * 17. The number of Sylow 3-subgroups of PH is congruent to 1 mod 3 and divides 17, so there is a
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Unformatted text preview: unique Sylow 3-subgroup which must be P , and we have P ± PH . Since groups of prime and prime squared order are abelian and P , H are normal subgroups of PH with coprime order, we can now deduce that PH is abelian. Similarly QH is abelian. Let h ∈ H . Then PH , QH ≤ C G ( h ) (because PH , QH are abelian), hence 3 2 * 17 and 5 2 * 17 divide | C G ( h ) | . Therefore 3825 ± ± | C G ( h ) and we deduce that C G ( h ) = G . We conclude that h ∈ Z ( G ) for all h ∈ H and it follows that H ≤ Z ( G ) ....
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## This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.

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