# Fundamental Theorem of Abelian Groups

1. Sep 28, 2011

### mehtamonica

Show that there are two abelian groups of order 108 that have exactly one subgroup of order 3.

108 = 2^ 2 X 3 ^ 3

Using the fundamental theorem of finite abelian groups, we have

Possible abelian groups of order 108 can be : Z108, Z4 + Z27, Z2+Z2+Z27, Z4+Z9+Z3, Z2+Z2+Z9+Z3, Z4+Z3+Z3+Z3, Z2+Z2+Z3+Z3+Z3

It seems to me that all three Z108, Z4+Z27, Z2+Z2+Z27 have exactly one subgroup of order 3. Please suggest where am I going wrong ?

2. Sep 28, 2011

### micromass

Staff Emeritus
The groups $\mathbb{Z}_{108}$ and $\mathbb{Z}_4\times \mathbb{Z}_{27}$ are isomorphic.

3. Sep 28, 2011

### mehtamonica

Thanks, Micromass.

4. Sep 28, 2011

### Bacle

Sorry to nitpick, mehtamonica: You may be using shorthand, but I think that should be finitely-generated Abelian groups, not finite Abelian groups.

5. Sep 28, 2011

### mathwonk

huh?

6. Sep 29, 2011

### Bacle

Well, in the OP, in line 3, Mehtamonica referred to the' fundamental theorem of

finite Abelian groups' ; Z -integers is clearly not finite; so it is the FT of fin.gen.

Abelian groups.

7. Sep 29, 2011

### Jamma

Yes, it is usually phrased as finitely generated, although of course it will imply as a corollary that all finite ones must be the product of torsion groups (because they can't contain a Z term, and will obviously all be finitely generated still).

8. Sep 30, 2011

### Bacle

The groups [\tex]Z108 , Z4×Z27 [tex] are isomorphic

Yes, but [\tex] Z108 [tex] has more music and less commercials!

Always thought the [\tex] Z100's [tex] sound like radio stations.

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