When, if ever, does $(I + H)/H \not\cong I/H$? (Equivalently, when $I/(H\cap I) \not\cong I/H$?)

Question

The quotient $G /H$ of two abelian groups, $H \subseteq G$, is defined as the set of all cosets of $H$: $$G/H := \{ g+H: g \in G \} \,. $$

Question: How does this definition depend on the fact that $H \subseteq G$? It seems like it only depends on addition of elements of $G$ and elements of $H$ being defined.

More specifically, let's say we have another subgroup of $G$, call it $I$, and possibly $H \not\subseteq I$.

1. Nevertheless, addition of elements of $H$ and $I$ is well-defined, so why is the set $$"I/H" := \{ g+H: g \in I \} $$ not well-defined as a group?

2. If it is well-defined as a group, why is not equal (isomorphic) to $(I+H)/H$ in general?

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Attempt: The fact that all groups are abelian simplifies things inasmuch as we are guaranteed that all subgroups are normal subgroups. But we still need to check in this case whether addition in $"I/H"$ is well-defined, since possibly $H \not\subseteq I$.

So let's say that $x_1, x_2 \in I$ are such that $x_1 + H = x_2 +H$, and $y_1, y_2 \in I$ are such that $y_1 +H = y_2 + H$. Then $(x_1 +H) + (y_1 +H) = (x_2 +H) + (y_2 +H)$ if and only if $$(x_1 + y_1) + H = (x_2 +y _2) + H \iff (x_1 + y_1) - (x_2 + y_2) \in H \,.$$ But $(x_1 + y_1) - (x_2 + y_2) = (x_1 - x_2) + (y_1 - y_2)$. And $(x_1 - x_2) \in H$ since $x_1 + H = x_2 + H$, and also $(y_1 - y_2) \in H$ since $y_1 +H = y_2 +H$, and since $H$ is a group, $(x_1 - x_2), (y_1 - y_2) \in H \implies (x_1 +y_1) - (x_2 + y_2) = (x_1 - x_2) + (y_1 - y_2) \in H$.

In other words, addition in $"I /H"$ does seem well-defined. Also, $"I/H"$ is closed under inverses, because $I$ is. And since $"I/H"$ is a subset of $G/H$ closed under addition and forming inverses, it should be a subgroup of $G/H$. Therefore, at no point does the assumption $H \subseteq I$ seem necessary.

Motivation: I started thinking about this when trying to show that the quotient $R/I$ of a graded ring $R$ by a homogeneous ideal $I$ is again a graded ring, because the first thing I wrote down for the possible grading of $R/I$ was $\oplus_{d \ge 0} R_d/I$, before realizing that, in general, $I \not\subseteq R_d$. This led me to try $\oplus_{d \ge 0} (R_d +I)/I$, which worked.

But now I think I may have been led to the right answer for the wrong reason, since I can't justify my original claim that $I \not\subseteq R_d$ implies that $R_d /I$ is not well-defined.

Note: This question is related, but is even more general, since it does not restrict to abelian groups. However, neither of the two answers to that question (one deleted) explain why the subset condition is necessary.

Answer 1

I've addressed the more general noncommutative version of your question in this answer, so here I'll address your motivation question directly:

The issue with "$R_d/I$", given that $I$ is not a subgroup of $R_d$, isn't one of well-definedness. It's just nonstandard terminology. The set of cosets of $I$ by elements of $R_d$ is, in fact, equal, both as a set and as a group, to $(R_d+I)/I$.

Moreover, $(R_d+I)/I$ is isomorphic to $R_d/(I \cap R_d)$, so that gives two ways to describe the group using standard notation. The fact that "$R_d/I$" is nonstandard is simply an emphasis of the fact that every "quotient" of $R_d$ can come from a normal subgroup of $R_d$.