How many subgroups of order p^{2 }does the abelian group the direct sum Z_{p3}+Z_{p2}?

I still don't have any idea.

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- Nov 11th 2012, 09:03 PMRitaSubgroups of order p^2 of an abelian group
How many subgroups of order p

^{2 }does the abelian group the direct sum Z_{p3}+Z_{p2}?

I still don't have any idea. - Nov 12th 2012, 01:39 AMDevenoRe: Subgroups of order p^2 of an abelian group
there are two types of groups of order p

^{2}: cyclic, and non-cyclic.

let's look at the non-cyclic groups of order p^{2}in Z_{p3}x Z_{p2}, first.

such a subgroup must be of the form HxK, where H is a subgroup of order p in Z_{p3}, and K is a subgroup of order p in Z_{p2}.

since there is only ONE such subgroup in each factor, we conclude the sole non-cyclic group of order p^{2}in Z_{p3}x Z_{p2}

is: <p^{2}> x <p>.

now if we have a CYCLIC subgroup K of order p^{2}of Z_{p3}x Z_{p2},

this means we have K = <(a,b)>, so either a is of order p^{2}(and b is arbitrary), or b is of order p^{2}and a is of order 1 or p.

there are p^{3}-p^{2}elements of order p^{2}in Z_{p3}, and p^{2}-p elements of order p^{2}in Z_{p2}.

this means we have a total of p^{4}-p^{3}+ p^{3}-p^{2}= p^{4}-p^{2}elements of order p^{2}in Z_{p3}x Z_{p2}.

since each distinct subgroup accounts for p^{2}-p of these elements (a cyclic group of order p^{2}has p^{2}-p elements of order p^{2}), we must have:

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

cyclic subgroups of order p^{2}.

this gives, by my accounting, p^{2}+p+1 subgroups of order p^{2}in all.

let's test this for p = 3:

non-cyclic subgroups of order 9 in Z_{27}x Z_{3}:

{(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 Z_{27}, 9 and 18 (= -9).

and there are exactly two elements of order 3 in Z_{9}, 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 Z_{27}x Z_{9}. 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 Z_{27}x Z_{9}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 Z_{27}x Z_{9}).

together with the unique non-cyclic subgroup of order 9 exhibited above, that is 13 in all, which is 3^{2}+3+1.