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Math Help - Cyclic group question - from Herstien

  1. #1
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    Cyclic group question - from Herstien

    This is Q#9, Section 2.9, pg 75 from Herstien
    Q: order(G) = pq, where p and q are distinct primes. G has a normal subgroup of order p and a normal subgroup of order q. Prove G is cyclic.

    My attempt:

    Let N1 be the subgroup of order p
    Let N2 be the subgroup of order q

    Clearly, N1 and N2 are cyclic (as for every group with prime order). Let a, b be generator of N1 and N2 respectively.

    Consider c=ab

    I can show that order(c)=pq. And, hence G is cyclic (with c as generator)

    Help I need:
    1. Is my attempt correct?
    2. I never used the fact that N1 and N2 are normal (something mentioned in the question) - So am I missing / mis-understanding something?

    Thanks
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  2. #2
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    If you've never used the fact that N_1 and N_2 were normal subgroups:

    S_3 is a group of order 6, it contains an element of order 2 and another one of order 3, which of course generate subgroups of orders 2 and 3.
    But S_3 is not cyclic...

    The hypothesis " N_1,N_2 normal" allows you to prove \text{order}(ab)=pq, for instance, consider the commutator [a,b]=aba^{-1}b^{-1} and prove it is the identity element, that will mean a,b commute and then you will be able to conclude \text{order}(ab)=pq.
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  3. #3
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    Quote Originally Posted by clic-clac View Post
    If you've never used the fact that N_1 and N_2 were normal subgroups:

    S_3 is a group of order 6, it contains an element of order 2 and another one of order 3, which of course generate subgroups of orders 2 and 3.
    But S_3 is not cyclic...

    The hypothesis " N_1,N_2 normal" allows you to prove \text{order}(ab)=pq, for instance, consider the commutator [a,b]=aba^{-1}b^{-1} and prove it is the identity element, that will mean a,b commute and then you will be able to conclude \text{order}(ab)=pq.
    Absolutely. Thanks very much. I missed the fact that to prove order(ab)=pq, I am implicitly using the fact that the sub-groups are normal.
    Thanks again !!
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    This is Q#9, Section 2.9, pg 75 from Herstien
    Q: order(G) = pq, where p and q are distinct primes. G has a normal subgroup of order p and a normal subgroup of order q. Prove G is cyclic.

    My attempt:

    Let N1 be the subgroup of order p
    Let N2 be the subgroup of order q

    Clearly, N1 and N2 are cyclic (as for every group with prime order). Let a, b be generator of N1 and N2 respectively.

    Consider c=ab

    I can show that order(c)=pq. And, hence G is cyclic (with c as generator)

    Help I need:
    1. Is my attempt correct?
    2. I never used the fact that N1 and N2 are normal (something mentioned in the question) - So am I missing / mis-understanding something?

    Thanks
    Of course you used the fact that both sbgps. are normal (otherwise the claim is false: there are non-cyclic, and even non-abelian groups of order pq, when one of the primes divides (the other one minus one)) ,but you didn't pay attention: how did you prove ord(c)=pq??

    Tonio
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