Abelian Group Order: Find Largest Possible # of Subgroups

[Dawson64]

This isn't homework, it was proposed by a professor of mine and I'm dying here because the hint makes no sense to me

Let G be an abelian group of order 540, what is the largest possible number of subgroups of order 3 such a group G can have?
He said to classify abelian groups of order 27, but I'm not sure how that's related. Anyone know how to solve this?

 

[micromass]

The fundamental theorem of group theory says that any group G of order 540 can be expressed as

G=G_1\times G_2

where G1 is a group of order 27 and G2 is a group of order 20.

If your searching for every group of order 3, then you'll actually be search for every element of order 3. (number of subgroups of order 3= number of elements of order 3 divides by 2). Since an element of order 3 can only be in G1, you'll be searching for elements of order 3 in a group of order 27. This will require classifying all groups of order 27...

 

[Dawson64]

The fundamental theorem of group theory says that any group G of order 540 can be expressed as

G=G_1\times G_2

where G1 is a group of order 27 and G2 is a group of order 20.

If your searching for every group of order 3, then you'll actually be search for every element of order 3. (number of subgroups of order 3= number of elements of order 3 divides by 2). Since an element of order 3 can only be in G1, you'll be searching for elements of order 3 in a group of order 27. This will require classifying all groups of order 27...

Is this an alright proof?

First note that Z(n) has exactly one d-subgroup, where d is any divisor of n.
Using the fundamental theorem of finite abelian groups, G can be isomorphic to either:
Z540, which has only one 3-subgroup,
Z3 x Z180, which has three 3-subgroups (namely, the subgroups generated by (1,0), (0, 60), and (1, 60))
Z3 x Z3 x Z60, which has seven (generated by (1,0,0), (0,1,0), (0,0,60), (1,1,0), (1,0,60), (0,1,60), and (1,1,60)), or
Z3 x Z3 x Z3 x Z20, which also has seven.

Note that there other possibilities to which G may be isomorphic, but you can check and see that none of them will yield more than seven 3-subgroups.

So the answer is seven.

And doesn't Z27 only have one 3-subgroup?

 

[micromass]

Yes, this seems to be correct.

Please note tho, that

Z3 x Z3 x Z60 = Z3 x Z3 x Z3 x Z20

 

[Dawson64]

Lol, whoops.