Math4302 Modern Algebra (Lecture 23)
Group
Group acting on a set
Theorem for the orbit of a set with prime power group
Suppose is a -set, and where is prime, then .
Where
Corollary: Cauchy’s theorem
If , where is a prime, divides , then has a subgroup of order . (equivalently, has an element of order )
This does not hold when is not prime.
Consider with order , and has no subgroup of order .
Corollary: Center of prime power group is non-trivial
If , then is non-trivial. ()
Proof
Let act on via conjugation, then . This makes to a -set.
Apply the theorem, the set of elements with trivial orbit is; Let , then .
Therefore .
So divides , so , therefore is non-trivial.
Proposition: Prime square group is abelian
If , where is a prime, then is abelian.
Proof
Since is a subgroup of , divides so or .
By corollary center of prime power group is non-trivial, .
If . If , then consider the group (Note that ). We have so is cyclic (by problem 13.39), therefore is abelian.
If , then is abelian.
Classification of small order
Let be a group
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- (prime order)
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- (prime order)
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- (prime order)
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Proof
has an element of order , namely , and an element of order , namely .
So are distinct.
Therefore, there are only two possibilities for value of . ( are inverse of each other, is inverse of itself.)
If , then is abelian, then .
If , then .
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- (prime order)
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- quaternion group where , . , , , , , .
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- (apply the corollary, , these are all the possible cases)
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- (prime order)
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- ??? One more
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- (prime order)
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Lemma for group of order where is prime
If is prime, , and , then is either abelian or
Proof
We know has an element of order 2, namely , and an element of order , namely .
So are distinct elements of .
Consider , if , then is abelian, then .
If , then .
cannot be inverse of other elements, if , where , then , then , then , therefore , then , so , therefore or .
This is not possible since .