# Thread: How Many Abelian Groups of Given Order

1. ## How Many Abelian Groups of Given Order

Up to isomorphism, how many additive abelian groups G of order 16 have the property that x+x+x+x=0 for each x in G?

I don't know where to start with this problem. I don't know how to figure how many additive abelian groups of order 16 there are to even begin narrowing it down to those with the given property. What property limits the number of abelian groups?

2. Originally Posted by StarKid
Up to isomorphism, how many additive abelian groups G of order 16 have the property that x+x+x+x=0 for each x in G?

I don't know where to start with this problem. I don't know how to figure how many additive abelian groups of order 16 there are to even begin narrowing it down to those with the given property. What property limits the number of abelian groups?
The only abelian groups up to isomorphism of order $16$ are $\mathbb{Z}_{16},\mathbb{Z}_8\oplus\mathbb{Z}_2,\ma thbb{Z}_4\oplus\mathbb{Z}_4,\mathbb{Z}_4\oplus\mat hbb{Z}_2\oplus\mathbb{Z}_$ and $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2\o plus\mathbb{Z}_2$. Evidently a necessary and sufficient condition for $4x=0$ is that $4x_j=0$ for each of the coordinates $x_j$ of $x$. But, of course it is necessary to check that this is true for the generators of each of the coordinate groups. Thus, it's clear with this formulat is that there is 3 groups, the last three I listed.

3. the thing is, abelian groups factor nicely into direct sums, in much the same way as vector spaces factor nicely into subspaces. so you only have to look for ways 16 can be factored as an integer (this is a consequence of the classification theorem for finite abelian groups, which are very nice groups to work with):

1*16 --> this gives you Z16
2*8 --> this gives you Z8⊕Z2 (same as, well isomorphic to, Z2⊕Z8)
4*4 --> this gives you Z4⊕Z4
2*2*4 ---> this gives you Z4⊕Z2⊕Z2
2*2*2*2 ---> this gives you Z2⊕Z2⊕Z2⊕Z2

the first two are out of the question, as 1 is of order 16 in Z16, and (1,0) is of order 8 in Z8⊕Z2.

but for (a,b) in Z4⊕Z4 it's clear that (4a,4ab) = (0,0), similarly for (a,b,c) in Z4⊕Z2⊕Z2 it's clear that (4a,4b,4c) = (0,0,0), and for (a,b,c,d) in Z2⊕Z2⊕Z2,

(4a,4b,4c,4d) = 2(2a,2b,2c,2d) = 2(0,0,0,0) = (0,0,0,0), so they all qualify.