# FiniteAbelian - Finite Abelian Groups 1 Preliminaries...

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Finite Abelian Groups 1 Preliminaries Defnition 1. Let H 1 ,H 2 ,...,H k be subgroups of G . We say that G is the internal direct product of H 1 k if the following hold. 1. H i ± G , for all H i 2. G = H 1 H 2 ··· H k 3. H i H 1 ± H i H k = { e } ( H 1 ± H i H k means the product of all the H ’s except H i ) Lemma 2. G be the internal direct product of H 1 k . Then we have the following. 1. H i H j = { e } for all i ± = j . 2. If g G , then there exists unique elements h i H i such that g = h 1 h 2 h k . 3. If h i H i and h j H j , then h i h j = h j h i . Proof. Exercise. Theorem 3. If G is the internal direct product of H 1 k , then G = H 1 ×··· H k . Proof. “Sketch” Defne φ : H 1 ×···× H k G by φ ( h 1 ,...,h k )= h 1 h k . Since G = H 1 H k , we get onto. By lemma 2 part 2 we get one-to-one, and fnally by lemma 2 part 3 we get that φ is a homomorphism. 2 Fundamental Theorem of Finite Abelian Groups Lemma 4. G be an abelian group such that | G | = p k a where p is prime and ( p, a )=1 . Then there exists a unique subgroup of order p k . Proof. By corollary 2.78 (in the revised printing) we know that such a subgroup exists. So let H G such that | H | = p k , and let G ( p { x G | x p k = e } . By problem #4 we know G ( p ) G . We claim that H = G ( p ). Let h H , then h p k = e since | H | = p k . Hence h G ( p ), so H G ( p ). Now let g G ( p ). So g p k = e . Consider gH G/H . Since | G/H | = a , ( ) | a . However ( ) |◦ ( g ), so ( ) | p k .S o ( ) = 1 since ( a, p ) = 1. ThereFore = H and hence g H . Thus G ( p ) H and hence H = G ( p ). ThereFore G ( p ) is the unique subgroup oF G oF order p k . Defnition 5. G be a group such that | G | = p k a where p is prime and ( p, a . A subgroup of order p k is called a Sylow p -subgroup of G .

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Theorem 6. Let G be a fnite abelian group such that | G | = p k 1 1 ··· p k l l where each oF the p i are distinct primes. Then G = G ( p 1 ) ×···× G ( p l ) . ProoF. We will show that G is the internal direct product of its Sylow p i subgroups and therefore by Theorem 3 we get our desired result. Firstly, since G is abelian all subgroups are normal, so G ( p i ) ± G . Next we will show that G = G ( p 1 ) G ( p 2 ) G ( p l ). Clearly G ( p 1 ) G ( p 2 ) G ( p l ) G . More- over, by homework problem # 3, we have that | G ( p 1 ) G ( p 2 ) G ( p l ) | = | G ( p 1 ) || G ( p 2 ) |···| G ( p l ) | = p k 1 1 p k 2 2 p k l l = | G | . Hence G = G ( p 1 ) G ( p 2 ) G ( p l ). Lastly we need to show that G ( p i ) G ( p 1 ) ± G ( p i ) G ( p l )= { e } for all i ∈{ 1 , 2 ,...,k } . Fix i 1 , 2 } . Then we can write the order of G as | G | = p k i i a where ( p i ,a ) = 1. Again by homework problem # 3, | G ( p 1 ) ± G ( p i ) G ( p l ) | = a . However, | G ( p i ) | = p k i . Hence ± | G ( p 1 ) ± G ( p i ) G ( p l ) | , | G ( p i ) | ² = 1 and thus G ( p i ) G ( p 1 ) ± G ( p i ) G ( p l { e } . Therefore G is the internal product of G ( p 1 ) ,...,G ( p l ) and hence G is isomorphic to G ( p 1 ) G ( p l ). Note that the above theorem proves that G is isomorphic to the direct product of its Sylow p i - subgroups. We will now show that each of these Sylow p i -subgroups can be decomposed into cyclic subgroups. For the following theorem we will be working with a p -group. So | G | = p n . Therefore all elements in G must have order p j for some j Z . Thus G contains an element of maximal order. That is, there exists a
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## This note was uploaded on 01/18/2012 for the course INFORMATIK 2011 taught by Professor Phanthuongcang during the Winter '11 term at Cornell University (Engineering School).

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FiniteAbelian - Finite Abelian Groups 1 Preliminaries...

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