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Posted by GiGi 3 years ago In:

GiGi
GiGi Asked:

I need to prove that no group of order 24 is simple using the Sylow Theorems. Also same question for order 56 and order 105. Please help.

I need to prove that no group of order 24 is simple using the Sylow Theorems.

Also same question for order 56 and order 105.

Please help.

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Answers

Ian
Ian Answered:

a) Prove that no group of order 24 is simple using the Sylow Theorems.
b) and c) Same question for order 56 and order 105.
Never heard of them, but tried WIKI and another website found these.
I hope you know what it means - I do not, but think you might.
a) Small groups are not simple
This example involves the order of the smallest simple group which is not cyclic.
Burnside's p^a*q^b theorem states that if the order of a group is the product of two prime powers, then it is solvable, and so the group is not simple, or is of prime order and is cyclic.
This rules out every group up to order 30 (= 2 · 3 · 5) and so rules out 24 it seems.
b) and c) From # 22 posting at this website
http://www.mathisfunforum.com/viewtopic.php?id=10734

Groups of order 30 56 105


G| 30   It then has 1 or 6 Sylow 5-subgroups and 1 or 10 Sylow 3-subgroups, each Sylow subgroup being cyclic of prime order. If there are 6 Sylow 5-subgroups, the union of these subgroups has elements; if there are 10 Sylow 3-subgroups, their union will have elements, all distinct from the elements of the other union except the identity. Since is only 30, therefore cannot have 6 Sylow 5-subgroups and 10 Sylow 3-subgroups at the same time; hence must have either a unique (therefore normal) Sylow 5-subgroup or a unique (normal) Sylow 3-subgroup.
G| 56   It then has 1 or 15 Sylow 7-subgroups and 1 or 21 Sylow 5-subgroups (each Sylow subgroup being cyclic of prime order). By a similar argument to the above, we find that must have either a normal Sylow 7-subgroup or a normal Sylow 5-subgroup.
G| 105   It then has 1 or 8 Sylow 7-subgroups (cylic of order 7). If 8, then the union of the the Sylow 7-subgroups has 48 non-identity elements. This leaves 56 - 48 = 8 other elements in the group (including the identity), which must then make up the unique Sylow 2-subgroup. Hence must have either a normal Sylow 7-subgroup or a normal Sylow 2-subgroup.

Hope it is useful, (I have no idea about this),
Regards - Ian
 

 

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