Partitioning a Group Into Disjoint Subgroups

[metapuff]

Hey everyone, I've got a question in elementary group theory.

Suppose we have a group G, and we want to completely partition it into multiple subgroups, such that the only element each subgroup shares with any other is the identity element. Is this ever possible?

I think that such a partitioning is impossible for any cyclic group. If g is a generator for a cyclic group G, and H is a subgroup of G with a smaller order than G, then g $\notin$ H, because closure of H would require all of g's powers (and hence all of G) to be in H. Therefore we cannot include the generators of G in any of the partitions, so partitioning is impossible.

What about non-cyclic groups, like the Klein Four Group?

Thanks!

 

[Terandol]

You really answered your own question already. You suggested the Klein four group so just check it. It is trivial to verify that the nontrivial subgroups of $\mathbb{Z_2}\times\mathbb{Z_2}=\{ (0,0),(1,0),(0,1),(1,1) \}$ are $H_1=\{(0,0),(1,0) \}, H_2=\{(0,0),(0,1) \}$ and $H_3=\{(0,0),(1,1) \}$. Further these subgroups intersect trivially and every element of the Klein four group is in one of them so they satisfy your conditions.

 

[metapuff]

Ah, of course! You're quite right. I wonder if all non-cyclic groups can be partitioned as such? Again, thanks for reminding me of what should have been obvious.

 

[micromass]

Ah, of course! You're quite right. I wonder if all non-cyclic groups can be partitioned as such? Again, thanks for reminding me of what should have been obvious.

Here's a hint. Let $G$ be a group and let $H$ be one of the partitioning sets. If $g\in H$, then $\langle g \rangle\subseteq := \{g^m~\vert~m\in \mathbb{Z}\} \subseteq H$. Thus the partitioning sets consists out of such "cycles". Now, you can visually represent such cycles by a cycle graph: http://en.wikipedia.org/wiki/Cycle_graph_(algebra )

Can you use the cycle graphs on the above wiki page and on this page: http://en.wikipedia.org/wiki/List_of_small_groups to test your conjecture?

 

[metapuff]

Ah, I hadn't seen cycle graphs before. It looks like we can partition ANY non-cyclic group into disjoint subgroups! Any group who's cycle graph has more than one closed loop can be partitioned as such, and any non-cyclic group has to have more than one closed loop in its cycle graph.

Although, hold on. Maybe I'm being too hasty. Just looking at the cycle graph for the Quaternions, we can see that all subgroups of the Quaternions need to have -1 in them, so we can't partition the Quaternions, even though they aren't cyclic. I'll need to think about this a bit more. In any case, thanks guys!

 

[micromass]

Ah, I hadn't seen cycle graphs before. It looks like we can partition ANY non-cyclic group into disjoint subgroups! Any group who's cycle graph has more than one closed loop can be partitioned as such, and any non-cyclic group has to have more than one closed loop in its cycle graph.

Although, hold on. Maybe I'm being too hasty. Just looking at the cycle graph for the Quaternions, we can see that all subgroups of the Quaternions need to have -1 in them, so we can't partition the Quaternions, even though they aren't cyclic. I'll need to think about this a bit more. In any case, thanks guys!

Indeed, the quaternions form a counterexample, unless you allow a partition with one subgroup.

 

[Incnis Mrsi]

For some groups it is possible. Example with Klein group can be easily generalized to vector spaces (of dimension > 1) over an arbitrary field, as groups under addition: they can be partitioned to one-dimensional subspaces parametrized by the projectivization of that vector space. Five years ago Ī even uploaded several nice pictures for finite fields. You can also generalize it to free modules over Euclidean domains: for instance, ℤ² admits a partition, by the same principle, to infinite number of subgroups isomorphic to ℤ.