Thread: Subgroups of order p^2 of an abelian group

How many subgroups of order p² 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 p²: cyclic, and non-cyclic.

let's look at the non-cyclic groups of order p² 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 p² in Z_p³ x Z_p²

is: <p²> x <p>.

now if we have a CYCLIC subgroup K of order p² of Z_p³ x Z_p²,

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 Z_p³, and p²-p elements of order p² in Z_p².

this means we have a total of p⁴-p³ + p³-p² = p⁴-p² elements of order p² in Z_p³ x Z_p².

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:



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.