Sylow p-Subgroups, Conjugacy and Abelian Groups
A) Let G be a group of order 203. Prove that if H is normal subgroup of order 7 in G then H<=Z(G). Deduce that G is abelian in this case.
b)Let P be a normal Sylow p-subgroup of G and let H be any subgroup of G. Prove that P intersect H is the unique Sylow p-subgroup of H.
c)Let P be in Syl_p(G) and assume N is a normal subgroup of G. Use the conjugacy part of Sylow's theorem to prove that P intesect N is a Sylow p-subgroup of N. Deduce that PN/N is a Sylow p-subgroup of G/n
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1. Proof:
We know , 7 and 29 are prime numbers. According to the third Sylow Theorem, if is the number of Sylow-29 subgroups in , then (mod 29). Thus . We know a group of prime order must be a cyclic group, so every two distinct groups with order 29 can only intersects at identity element. So the only possibility is . This means that has a unique Sylow-29 subgroups and thus is a normal subgroup.
According to the condition, the sylow-7 ...
Solution Summary
Sylow p-Subgroups, Conjugacy and Abelian Groups are investigated. The solution is detailed and well presented.