Fundamental Theorem of Abelian Groups

[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 ?

 

[micromass]

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 ?

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

 

[mehtamonica]

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

Thanks, Micromass.

 

[Bacle]

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

 

[mathwonk]

huh?

 

[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.

 

[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).

 

[Bacle]

The groups [\tex]Z108 , Z4×Z27 are isomorphic<br /> <br /> Yes, but [\tex] Z108 has more music and less commercials!&lt;br /&gt; &lt;br /&gt; Always thought the [\tex] Z100&amp;#039;s sound like radio stations.