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Math Help - Finite Abelian Group

  1. #1
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    Finite Abelian Group

    Question: G is a finite abelian group of order n. Let d|n.
    Prove G has a subgroup of order 'd'.

    I know that if p^a (where p is a prime and a any +ve integer) divides n, then G has a subgroup of order p^a

    I was trying to use this for the question. Is this correct approach? I have not read the generic theory of abelian groups, so would want to avoid that as much.

    Any help/pointers please?

    Thanks
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    Question: G is a finite abelian group of order n. Let d|n.
    Prove G has a subgroup of order 'd'.

    I know that if p^a (where p is a prime and a any +ve integer) divides n, then G has a subgroup of order p^a

    I was trying to use this for the question. Is this correct approach? I have not read the generic theory of abelian groups, so would want to avoid that as much.

    Any help/pointers please?

    Thanks

    Of course this is enough, but you must know:

    i) If H,K are two sbgps. of a group G, then HK is again a sbgp. iff HK = KH

    ii) If N is a NORMAL sbgp. of a group G, then NK is always a sbgp. of G for any sbgp. K of G

    iii) For any two sbgps. of an ABELIAN group G, HK is again a sbgp. of G.

    iv) For H,K sbgps. of G, |HK| = |H||K|/|H /\ K|

    Take some minutes to fully understand the above and then try to put things together.
    Note also that (i ==> (ii) ==> (iii)

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Of course this is enough, but you must know:

    i) If H,K are two sbgps. of a group G, then HK is again a sbgp. iff HK = KH

    ii) If N is a NORMAL sbgp. of a group G, then NK is always a sbgp. of G for any sbgp. K of G

    iii) For any two sbgps. of an ABELIAN group G, HK is again a sbgp. of G.

    iv) For H,K sbgps. of G, |HK| = |H||K|/|H /\ K|

    Take some minutes to fully understand the above and then try to put things together.
    Note also that (i ==> (ii) ==> (iii)

    Tonio
    Thanks Tonio. Will read what all you wrote carefully and attempt it.
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  4. #4
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    Quote Originally Posted by tonio View Post
    Of course this is enough, but you must know:

    i) If H,K are two sbgps. of a group G, then HK is again a sbgp. iff HK = KH

    ii) If N is a NORMAL sbgp. of a group G, then NK is always a sbgp. of G for any sbgp. K of G

    iii) For any two sbgps. of an ABELIAN group G, HK is again a sbgp. of G.

    iv) For H,K sbgps. of G, |HK| = |H||K|/|H /\ K|

    Take some minutes to fully understand the above and then try to put things together.
    Note also that (i ==> (ii) ==> (iii)

    Tonio

    Thanks a lot ! I think I got it. Let me write it and you can plz check.

    I understand - "Note also that (i ==> (ii) ==> (iii)" In an abelian G every sub-group is normal.


    Right, so here is my approach

    1. Prime Factorization of d - Let d = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}

    2. So we have sub-gorups H_1, H_2....H_k of order p_1^{a_1},p_2^{a_2},.....,p_k^{a_k} respectively

    3. Claim H = H_1H_2....H_k is the desired sub-group of order = d
    a. That it is a subgorup is obvious from your point (iii)
    b. To prove  |H| =  p_1^{a_1}p_2^{a_2}.....p_k^{a_k}, I will use induction.
    Basic logic of induction
    Consider H=H_1H_2
    H_1\cap H_2 is a sub-group of both H_1 and H_2. But gcd(H_1,H_2)=1, thus |H_1\cap H_2| =1
    Hence the result by your point# (iv)

    I think I have got it. If you could plz let me know if I missed/mistaked somewhere.

    And, thanks for your help.
    Last edited by aman_cc; October 17th 2009 at 04:55 AM.
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  5. #5
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    Quote Originally Posted by aman_cc View Post
    Thanks a lot ! I think I got it. Let me write it and you can plz check.

    I understand - "Note also that (i ==> (ii) ==> (iii)" In an abelian G every sub-group is normal.


    Right, so here is my approach

    1. Prime Factorization of d - Let d = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}

    2. So we have sub-gorups H_1, H_2....H_k of order p_1^{a_1},p_2^{a_2},.....,p_k^{a_k} respectively

    3. Claim H = H_1H_2....H_k is the desired sub-group of order = d
    a. That it is a subgorup is obvious from your point (iii)
    b. To prove  |H| = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}, I will use induction.
    Basic logic of induction
    Consider H=H_1H_2
    H_1\cap H_2 is a sub-group of both H_1 and H_2. But gcd(H_1,H_2)=1, thus |H_1\cap H_2| =1
    Hence the result by your point# (iv)

    I think I have got it. If you could plz let me know if I missed/mistaked somewhere.

    And, thanks for your help.

    Great! I'd give full marks for such a response.

    Tonio
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