Prove that a group of order 108 must have a normal subgroup of order 9 or 27.© BrainMass Inc. brainmass.com October 9, 2019, 6:34 pm ad1c9bdddf
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Group Theory (CXIII)
Groups of Order having Power of a Prime
Another Counting Principle
By:- Thokchom Sarojkumar Sinha
Prove that a group of order 108 must have a normal subgroup of order 9 or 27.
Solution:- Let be a group of order 108 . Then
The number of 3-Sylow subgroups of is of the form and
which imply that
If then there exists only one 3-sylow subgroup of , which is of order 27 and hence it is normal in .
If , then then there exists ...
It is proven that a group of order 108 must have a normal subgroup of order 9 or 27. The solution is detailed and well presented.