Let A be a nonzero additive abelian group.
We say A is decomposable provided there exists groups B and C, both nonzero, such that $A\simeq B\times C$.
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 two nonzero subgroups must also be nonzero.
Please, help.

Question

Let A be a nonzero additive abelian group.
We say A is decomposable provided there exists groups B and C, both nonzero, such that $A\simeq B\times C$.
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 two nonzero subgroups must also be nonzero. 
Please, help.

loser66 · Answer

@myininaya

anonymous · Answer

why not @loser69

fibonaccichick666 · Answer

ask @zepdrix ,because I do not know

myininaya · Answer

I have a lot of terms to look up.

myininaya · Answer

A is isomorphic to Cartesian product of B and C?

loser66 · Answer

yup

loser66 · Answer

Question: in additive abelian group, is it true that : if H1, H2, .... Hn are subgroup of G, then G = H1+H2 +...+Hn ?

loser66 · Answer

when Hi is indecomposable. 
For example : subgroups of Z , + are A = 2Z, B = 3Z, then Z = A+B

myininaya · Answer

I don't know...I can't recall that theorem or even know if I know how to prove it...
so we want to suppose A is an abelian nonzero group and that it is uniform...And we want to show B intersect C is nonzero given B and C are nonzero subgroups of A.
So A is a uniform abelian nonzero group...
This means what exactly...
Well are given that B and C are indecomposable.

loser66 · Answer

I got stuck at this problem because I see the conflict between the problems. I have another question like: Prove that the additive group Z is uniform. To me, that means (Z,+) is uniform, then, the intersection of any of 2 subgroups must be non zero, 
but obviously $2Z\cap3Z=\{0\}$

loser66 · Answer

So, how (Z,+) is uniform?

myininaya · Answer

This problem makes me cry. I can't look at it anymore. I'm sorry @loser66 .

loser66 · Answer

It's ok, friend. I do appreciate what you do for me.

myininaya · Answer

I think Zarkon could be more help on this.

myininaya · Answer

I honestly try to do a little research before I gave up...

myininaya · Answer

Just researching some of the definitions...And when I looked at all of it...It was like written in a different language.

loser66 · Answer

:) Ok, let me ask others for help. Thanks a lot.

anonymous · Answer

hmm i have no idea, but i am sure that 
$$3\mathbb{Z}\cap 2\mathbb{Z}\neq \{0\}$$

loser66 · Answer

how?

loser66 · Answer

$$3\mathbb Z=\{multiple~of~3\}$$
while $$2\mathbb Z=\{multiple~of~2\}$$

anonymous · Answer

6

loser66 · Answer

yes! :)

anonymous · Answer

So, the intersection isn't trivial.

loser66 · Answer

but we have a bunch of subgroups of Z, check each pair?

anonymous · Answer

no check $$n\mathbb{Z}\cap m\mathbb{Z}$$

loser66 · Answer

where n is odd and m is even, right? general case only, right?

anonymous · Answer

still have no idea how to answer your original question

anonymous · Answer

you are thinking too hard
doesn't matter what m and n are
maybe you are getting them confused with $$\mathbb{Z}_n$$

loser66 · Answer

nope, intersection of Z_n is always non trivial

anonymous · Answer

the answer lies in god

loser66 · Answer

I got it. Thanks all