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Math Help - Subgroups of order p^2 of an abelian group

  1. #1
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    Subgroups of order p^2 of an abelian group

    How many subgroups of order p2 does the abelian group the direct sum Zp3+Zp2?
    I still don't have any idea.
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  2. #2
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    Re: Subgroups of order p^2 of an abelian group

    there are two types of groups of order p2: cyclic, and non-cyclic.

    let's look at the non-cyclic groups of order p2 in Zp3 x Zp2, first.

    such a subgroup must be of the form HxK, where H is a subgroup of order p in Zp3, and K is a subgroup of order p in Zp2.

    since there is only ONE such subgroup in each factor, we conclude the sole non-cyclic group of order p2 in Zp3 x Zp2

    is: <p2> x <p>.

    now if we have a CYCLIC subgroup K of order p2 of Zp3 x Zp2,

    this means we have K = <(a,b)>, so either a is of order p2 (and b is arbitrary), or b is of order p2 and a is of order 1 or p.

    there are p3-p2 elements of order p2 in Zp3, and p2-p elements of order p2 in Zp2.

    this means we have a total of p4-p3 + p3-p2 = p4-p2 elements of order p2 in Zp3 x Zp2.

    since each distinct subgroup accounts for p2-p of these elements (a cyclic group of order p2 has p2-p elements of order p2), we must have:

    \frac{p^4 - p^2}{p^2 - p} = \frac{p^2(p^2 - 1)}{p(p-1)} = p^2+p

    cyclic subgroups of order p2.

    this gives, by my accounting, p2+p+1 subgroups of order p2 in all.

    let's test this for p = 3:

    non-cyclic subgroups of order 9 in Z27 x Z3:

    {(0,0),(9,0),(18,0),(0,3),(9,3),(18,3),(0,6),(9,6) ,(18,6)}

    are there any others? well there are exactly two elements of order 3 in Z27, 9 and 18 (= -9).

    and there are exactly two elements of order 3 in Z9, 3 and 6 (= -3).

    so the possible subgroups we might form are:

    <9> x <3>
    <18> x <3>
    <9> x <6>
    <18> x <6>

    but all of these yield the group listed above, there are no others.

    now, by our formula, we should find 12 different cyclic subgroups of order 9 in Z27 x Z9. let's see if we can list them:

    <(3,0)>
    <(3,1)>
    <(3,2)>
    <(3,3)>
    <(3,4)>
    <(3,5)>
    <(3,6)>
    <(3,7)>
    <(3,8)> that's 9 so far. convince yourself that we haven't yet duplicated any elements of order 9 (the group Z27 x Z9 has 243 elements, of which:

    162 are of order 27 -(a,b) with a of order 27, and arbitrary b
    72 are of order 9 - (a,b) with a of order 9, and b arbitrary (54), or a of order 3, and b of order 9 (12), or a = 0, and b of order 9 (6)
    8 are of order 3 - (a,b) with a of order 3, and b of order 3 (4), or a = 0 and b of order 3 (2), or a of order 3 and b = 0 (2)
    1 is of order 1 - (0,0)

    243 = 162+72+8+1)


    <(9,1)> that's 10.
    <(18,1)> it may not be obvious that this is different from <(9,1)> let's see:

    <(9,1)> = {(0,0),(9,1),(18,2),(0,3),(9,4),(18,5),(0,6),(9,7) ,(18,8)}

    <(18,1)> = {(0,0),(18,1),(9,2),(0,3),(18,4),(9,5),(0,6),(18,7 ),(9,8)} <---note no overlap of elements of order 9 (we do have overlap in the elements of order 3). 11 so far.

    and finally, <(0,1)> makes 12 (which accounts for ALL elements of order 9 in Z27 x Z9).

    together with the unique non-cyclic subgroup of order 9 exhibited above, that is 13 in all, which is 32+3+1.
    Last edited by Deveno; November 12th 2012 at 01:53 AM.
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