Sylow 5-subgroup

Related BrainMass Solutions

• Sylow's Theorem

  According to Sylow's Theorem, the number of its Sylow 5-subgroup is 5k+1 | 2^3 = 8. Then the only possibility is k = 0. So G has unique Sylow 5-subgroup and hence it must be normal.

• Prove that a group of order 120 is not simple.

  The number of Sylow 5-subgroup in G is 1 + 5k and 1 + 5k│24, then k = 0 , 1 If k = 0, then then Sylow 5-subgroup is normal in G.

• Sylow p-Subgroups, Conjugacy and Abelian Groups

  Suppose is a Sylow-p subgroup of , then is a p-subgroup of . According to the second Sylow Theorem, must be subgroup of a Sylow-p subgroup of .

• Subgroup proofs

  This provides an example of completing a proof about the largest normal subgroup in a group and Sylow 5-subgroups.

• Sylow's Theorem

  This solution is comprised of a detailed explanation to find a 2-Sylow subgroup and a 3-Sylow subgroup.

• Sylow's Theorem

  of or, Sylow 3-subgroups and 2-Sylow subgroup or, Sylow 2-subgroups of the symmetric group of degree 4, S4.

• Prove that a group of order 108 must have a normal subgroup of order 9 or 27.

  Otherwise , then there exists only one 3-Sylow subgroup of order 27 in and hence this 3-Sylow subgroup is normal in Hence a group of order 108 must have a normal subgroup of order 9 or, 27.

• Sylow Theorem Examined

  So G has unique Sylow q-subgroup. Still according to the Sylow Theorem, the Sylow q-subgroup is normal in G. Therefore, G is not simple. Sylow Theorem is contextualized.

• Application of Sylow's Theorems

  Therefore, if H is a group, then it is a Sylow-p subgroup of G.