Statement
Let the ring $R$ and its ideal $I$ and $J$ s.t. $I\subset J$
Then there is a subring(or ideal) $R_J(\simeq R/J)$ of the $R/I$
There are 2 questions.
First) I'm not sure the above statement is right. If the above statement is correct, How could I prove it?
Second) Let's substitute as the normal groups $N_1$, $N_2$ and $G$(group) instead of the ideals $I$,$J$ and $R$(ring).
Then does it true when we considering the group case?
First)
Not correct.
Surjective ringhomomorphism $\nu:R/I\to R/J$ prescribed by $r+I\mapsto r+J$ has kernel $J/I:=\{j+I\mid j\in J\}\subseteq R/I$.
Then $R/J$ and $(R/I)/(J/I)$ are isomorpic but $(R/I)/(J/I)$ is a quotient of $R/I$ is (not a subring or an ideal).
Second)
Yes, there is an analogy. Again we are dealing with a quotient then and not a subgroup.