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.
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:
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.