# Thread: [SOLVED] Show that G forms a ring.

1. ## [SOLVED] Show that G forms a ring.

Can someone look over what I have done and help me finish my proof? I have no idea how to prove distribution.
Thanks.

Exercise: Suppose G is an abelian group wrt addition and with identity 0. Define a multiplication in G by ab=0 for all a,b in G. Show that G forms a ring wrt these operations.

Since G is abelian wrt multiplication, I would like to use the last two conditions of the following definition.

Definition 5.1b. Alternative Definition of a Ring
Suppose R is a set in which a relation of equality and operation of addition and multiplication are defined. Then R is a ring wrt these operations if these conditions hold:
R forms an abelian group wrt addition.
R is closed wrt and associative multiplication.
Two distributive laws hold in R: x(y + z) = xy + xz, and (x + y)z = xz + yz for all x, y, and z in R.

Proof. Since G is an abelian group with respect to addition, then we only need to prove the last two conditions of the Alternative Definition of a Ring.
Let ab=0 and bc=0 for all a, b, c in G. This means that a(bc)=a(0)=0, and (ab)c=(0)a=0. This shows that G is closed under associative multiplication.
Now we need to show that the distributive laws hold.
...?

2. You were correct up to this point.

Originally Posted by yvonnehr
Let ab=0 and bc=0 for all a, b, c in G. This means that a(bc)=a(0)=0, and (ab)c=(0)a=0. This shows that G is closed under associative multiplication.
I think you mean: Let $a,b,c\in G.$ Then $ab=0\in G$ proving closure and $(ab)c=0c=0=a0=a(bc)$ proving associativity.

Distributivity is true because $0=0+0.$ Think about it.

3. Originally Posted by TheAbstractionist
You were correct up to this point.

I think you mean: Let $a,b,c\in G.$ Then $ab=0\in G$ proving closure and $(ab)c=0c=0=a0=a(bc)$ proving associativity.

Distributivity is true because $0=0+0.$ Think about it.
How did you get from 0 = a0 = a(bc)? I'm not sure if I missed an implication somewhere.

Thanks.

4. Originally Posted by yvonnehr
How did you get from 0 = a0 = a(bc)? I'm not sure if I missed an implication somewhere.

Thanks.
Remember that you let $ab=0=bc$...