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

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

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Modern Algebra

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

For

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 ...

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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.