Question

\(\huge Problem1\)
We say A is decomposable provided there exist groups B and C, both nonzero, such that A isomorphism to B x C. A nonzero additive abelian group that is not decomposable is said to be indecomposable. We say that A is "uniform" provided every nonzero subgroup of A is indecomposable.
Prove that a nonzero additive abelian group is uniform if and only if the intersection of any 2 nonzero subgroups must also be nonzero.
\(\huge Problem 2\)
1)Prove that the additive group Z is uniform.
2) Prove that the additive group Q is uniform.
3) Show that the additive group R is not uniform.


I do not know how to start.
Please help.

Answer 1

My understanding:
Let G be an additive abelian group, A1, A2, .....,As are subgroups of G. Since G is abelian, then \(A_i\) is normal.
By fundamental theorem of finite abelian group, \(G\simeq A_1\times A_2\times.....\times A_s\)
By defining above, each \(A_i\) is indecomposable.

Answer 2

We need show if \(A_i\cap A_j \neq \{0\}\) , then G is uniform.
and other direction is G is uniform, then \(A_i\cap A_j \neq \{0\}\)
But I don't see how to do. :(

Answer 3

@Alchemista

Answer 4

Give me example, please

Answer 5

Look at the definition of direct product.

Answer 6

I think of this \(G\simeq Z_2\times Z_3\) and both \(Z_2,Z_3\) are simple, and we know that simple groups are undecomposable. but \(Z_2\cap Z_3 \neq \{0\}\)

Answer 7

I am just down the problem into simple case to understand what is going on and find out how to argue in general.

Answer 8

But Z modulo is so special cases since every element of the set { Z_i} have the same elements. I mean \(Z_2 = \{[0],[1]\}\) , and \(Z_{whatever}=\{ start~from~[0],[1],....~~also\}\) so that intersection of them surely \(\neq \{0\}\)

Answer 9

Let's try to argue by contradiction.

Answer 10

Assume that intersection of any two nontrivial subgroups must also be nontrivial. By contradiction, let's suppose that \(A\) is not uniform. In other words there is a nontrivial proper subgroup of \(A\) which is decomposable.

Answer 11

I think the idea now is to show that if the intersection of every nontrivial subgroup is nontrivial then then the decomposable subgroup cannot exist.

Answer 12

Look at the section on direct products: http://math.stackexchange.com/questions/106028/semi-direct-v-s-direct-products

Answer 13

It requires the existence of \(H, K\) where \(H \cap K = \{e\}\) (the intersection is trivial).

Answer 14

We don't have it, right? It asks us to prove that the intersection is not identity.

Answer 15

Yes I think that's right. I have to check more carefully if it works though.