If $R$ is a ring and $U$ be the (two-way) ideal of a ring. Then we define, $R/U$ to be the set of all cosets of $R$ under addition. Prove that $R/U$ forms a group under addition defined in $R.$
I tried solving the problem as:
To prove that $R/U$ forms a group under addition, we need to show that the closure property, associative property, identity property and inverse property holds.
Let us denote $R/U=S.$
We show these properties one by one.
Closure Property: If $a+U,b+U\in S$ then, $$(a+U)+(b+U)=\color{green}{a+(U+b)+U}.$$ Now, as $b+U=U+b$ since, $R$ forms an abelian group under addition so, we may write this as, $a+b+U+U=a+b+U.$ Again, $a+b\in R$ implies $a+b+U\in S.$
Associative Property: If $a+U,b+U,c+U\in S,$ then $a+U+(b+U+c+U)=(a+U+b+U)+c+U.$ This is because, $R$ is an an abelian group under addition.
Identity Property: We note that $U+0=U\in S$ and this is precisely the identity element. For, all $a+U\in S$ we have, $a+U+U=U+a+U(=a+U+U)=a+U.$
Inverse Property: If $a+U\in S$ then, $-a\in R$ so, $-a+U\in S$ and we note that, $a+U-a+U=0+U=-a+U+a+U=U.$
This proves that $S$ is a group under addition.
However, there's a problem. I am not quite sure about the way, I have proved the closure property or not. This is because in the step $$(a+U)+(b+U)=\color{green}{a+(U+b)+U},$$ ain't I essentially using the associative property before even proving it that it holds?
My logic is simple. It is because that addition is associative in $R$ and that's the same addition operation's, working in here, so $(a+U)+(b+U)$ is the same as, $a+(U+b)+U$ and as addition's commutative as well so, I replaced $U+b$ by $b+U$ freely.
Another way to see is this, that if $u\in U$ is an arbitrary element, then $a+u+b+u=a+(u+b)+u$ as $a,b,u\in R$ and addition is associative in $R.$ This means $(a+U)+(b+U)=a+(U+b)+U$ is valid.
Also, as $\forall u,b\in R$ satisfying $u\in U$ we have, $u+b=b+u$ (since, $R$ is an abelian group under addition) so, $U+b=b+U.$
Using these two workouts above , we have a ready justification for the validity of the way of proving the closure and associative property in the proof.
Am I correct in my reasonings? Will the above proof, I presented can be improved more or can it be made for rigorous?
As pointed out in the comments, your proof is reasonable. In particular, your justification for "associativity" in the Closure Property part is valid.
Nitpicky things:
You should make sure you are following percisely your text's definition of a group. In modern treatments, the Closure Property is not included in the list of group axioms because it in some sense precedes them: when we define a binary operation as a function $\circ: G\times G\to G$, this Property is embedded in the implicit claim that $G$ is indeed a valid codomain for this function. In this writeup you have not defined $R/U$ as a set, so I assume this is done elsewhere, so you'll have to consult that as well to see if Closure was already done for you. (In any case, it was polite to your reader to write it out.)
Your "problem" in the 2nd section is also used in the Associativity Property proof, so your effort was not wasted either way :)
The proof of the Inverse Property needs a slight rewording: $a$ is not uniquely determined by $a+U$, so "choosing $-a$". Fortunately, choosing any element $x\in a+U$ we have $a+U=x+U$ and so we can simply take $-x$ and proceed as before. (Because of the structure of $S$ we can of course start with $a\in R$, but in my opinion this is a bad habit. We need to say things about any element in $S$, and so we "morally" should start the proof in $S$; for more complicated objects, this can save some logical headache.) Note this is a bit more subtle than what happens in the other parts: the Inverse Property is the only time where you have to conjure many "new" elements of $S$, whereas the others only need to "combine" old ones.
Typo in the Identity Property: the $U+0$ should be $0+U$.
As Arturo said, usually we call these "two-sided" ideals.