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Math Help - Abelian group, number of elements

  1. #1
    Senior Member slevvio's Avatar
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    Abelian group, number of elements

    Hello, I am stuck with the following question and was wondering if anyone could give me some advice.

    Let be an abelian group of order pq, where p<q are primes. Show that there are exactly p-1 elements of order q, q-1 elements of order q, and (p-1)(q-1) elements of order pq.

    OK so we have a theorem which tells us that there must be only one subgroup of order q (which is normal). This subgroup is cyclic because it is of prime order, so every non identity element is of order q. This gives q-1 elements of order q. There can be no other elements of order q in G since they would generate a distinct subgroup of order q, and there can be only one.

    So there are definitely only q-1 elements of order q in G.

    By Cauchy's theorem, there exists an element of order p in G. This generates a cyclic group of order p which has p-1 elements of order p. why can there be no more than that?
    Knowing this part implies the number of elements of order pq = pq - (p-1) - (q-1) = (p-1)(q-1), so how do I show that there are no more than p-1 elements of order p?

    Any help with this would be appreciated, Thank you
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  2. #2
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    Quote Originally Posted by slevvio View Post
    Hello, I am stuck with the following question and was wondering if anyone could give me some advice.

    Let be an abelian group of order pq, where p<q are primes. Show that there are exactly p-1 elements of order q, q-1 elements of order q, and (p-1)(q-1) elements of order pq.

    OK so we have a theorem which tells us that there must be only one subgroup of order q (which is normal). This subgroup is cyclic because it is of prime order, so every non identity element is of order q. This gives q-1 elements of order q. There can be no other elements of order q in G since they would generate a distinct subgroup of order q, and there can be only one.

    So there are definitely only q-1 elements of order q in G.

    By Cauchy's theorem, there exists an element of order p in G. This generates a cyclic group of order p which has p-1 elements of order p. why can there be no more than that?
    Knowing this part implies the number of elements of order pq = pq - (p-1) - (q-1) = (p-1)(q-1), so how do I show that there are no more than p-1 elements of order p?

    Any help with this would be appreciated, Thank you
    Hint: an abelian group of order pq, with p \neq q primes, is cyclic.
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  3. #3
    Senior Member slevvio's Avatar
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    I see why this would imply the result

    but why is the group cyclic ?
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  4. #4
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    Quote Originally Posted by slevvio View Post
    I see why this would imply the result

    but why is the group cyclic ?

    Because

    1) It is the direct product of its Sylow subgroups

    2) If C_r denotes the cyclic group of order r\in\mathbb{N} , then lcm(m,n)=1\Longrightarrow C_m\times C_n\cong C_{mn}

    Tonio
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