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Finite abelian group of size p-1

  1. Mar 14, 2009 #1
    Suppose we have two finite abelian groups [tex]G,G^{\prime}[/tex] of size [tex]n=pq[/tex], [tex]p,q[/tex] being primes. [tex]G[/tex] is cyclic.

    Both [tex]G,G^{\prime}[/tex] have subgroups [tex]H,H^{\prime}[/tex], both of size [tex]q[/tex]. The factor groups [tex]G/H,\ G^{\prime}/H^{\prime}[/tex] are cyclic and since they are of equal size, they are isomorphic. Are [tex]G,G^{\prime}[/tex] also isomorphic?

    Edit: The title is wrong. p-1 has nothing to do with this problem. Sorry about that.
     
    Last edited: Mar 14, 2009
  2. jcsd
  3. Mar 14, 2009 #2
    I don't think the groups G and G' are required to be isomorphic. You have already defined G so consider
    [tex] G' = <a,b|a^{p}=b^{q}=e> [/tex]
    where e is the identity. This should be a suitable counter example -try working through it.

    If I am wrong about it being a counter example then this should prove that G and G' are required to be isomorphic as there are only two non-isomorphic groups of order pq with p and q being prime.

    Hope this helps.
     
  4. Mar 15, 2009 #3
    EDIT: sorry for that earlier post

    Let p > q. Now, if q does not divide p-1 then any group of order p*q must be cyclic. (This is a result which can be proved easily with sylow's theorems)
    A simple example are groups of order 15 = 5*3.

    If on the other hand, q does divide p-1 then this group must be non-abelian. (The proof for this is a bit involved, but it can be found in some text book like herstein).

    Now since, you have already stated that G is cyclic. This implies q does not divide p-1. Then G must be cyclic and so must be G'. Hence they must be isomorphic.

    Suppose p = q. In this case, its really difficult to say anything, because G' has two possible structures:
    1. a cyclic group of order p^2.
    2. a group isomorphic to C_p x C_p. (C_p is the cyclic group of order p)

    I'll try to find the references containing the proofs for the results I have stated above and post them later, but I hope this was useful.
     
    Last edited: Mar 15, 2009
  5. Mar 15, 2009 #4

    matt grime

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    Thirsty dog, the group you defined for G' is infinite: it is the free product of C_p and C_q.


    Do you know an structure theorems for finite abelian groups? Are you supposed to assume that p and q are distinct primes?
     
  6. Mar 15, 2009 #5
    Hey Matt,

    I forgot to say Abelian... I am guessing you realised this by your next comment.

    I was under the impression that any finite Abelian group is expressible as the Cartesian products of a finite number of cyclic groups whose orders are prime powers.

    I think that if p and q are distinct then the cyclic group of order pq is isomorphic to the Cartesian of cyclic groups of order p and q. If p and q are equal then no such isomorphism exist.

    This says in terms of the original question G and G' are definitely isomorphic if p is distinct from q. But if p=q then G and G' are not always isomorphic. Ancient_Nomad has pointed this out.
     
  7. Mar 16, 2009 #6
    Intuitively, I would guess that H,H' are isomorphic => G,G' are isomorphic...
     
  8. Mar 16, 2009 #7
    Consider
    [tex] C_{4} = <c|c^{4}=e> \mbox{ and } C_{2}\times C_{2} = <a,b|a^{2}=b^{2}=e,ab=ba> [/tex]
    These are clearly not isomorphic as the first has an element of order 4 while the latter does not.

    We can choose the subgroups
    [tex] H = <c^{2}> \mbox{ and } H' = <a> [/tex]
    Thus H and H' are isomorphic. Also
    [tex] C_{4}/H \cong C_{2} \cong C_{2}\times C_{2}/H' [/tex]

    This shows that just because an two Abelian groups can have isomorphic subgroups and their quotients are also isomorphic doesn't imply the original groups are.
     
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