a problem about a generalization of cosets in a group.

Question

Let $f$ be a function defined a group $(A,+)$. Define $I(a)=\{x\in A \mid f(x)=f(a)\}$ for every $a\in A$. Assume that $I(a_0)=\{a_0\}$ where $a_0\in A$ and $a_0\notin I(a)$. In the following, $a$ and $a_0$ can be treated as constants. Can we say:

1. $2a_0 \in C$ implies $I(a)-a_0=a_0-I(a)$?

2. either $I(a)-a_0=a_0-I(a)$ or $(I(a)-a_0)\cap (a_0-I(a))=\emptyset$ holds?

where $C=\{a+b \mid a\in I(a), b\in I(a)\}$, $I(a)-a_0=\{b-a_0\mid b\in I(a)\}$ and $a_0-I(a)=\{a_0-b\mid b\in I(a)\}$.

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If $I(a)$ is a subgroup of $A$ and $a_0=0$, the answer to the two questions are YES.

My Attempt:

1. For the first question: Since $2a_0\in C$, there exist $a_1,a_2\in I(a)$ such that $2a_0=a_1+a_2$ and $f(a_1)=f(a_2)=f(a)$. Now for every $x\in I(a)-a_0$, there exists $a_3\in I(a)$ such that $x=a_3-a_0$ and $f(a_3)=f(a)$. We have $$x=a_3-a_0=a_0+a_3-2a_0=a_0+a_3-a_1-a_2=a_0-(a_1+a_2-a_3).$$ I am stuck here. It is hard to say $a_1+a_2-a_3\in I(a)$.

2. For the second question: If $(I(a)-a_0)\cap (a_0-I(a))\ne \emptyset$, then suppose that $x\in (I(a)-a_0)\cap (a_0-I(a))$. There exist $a_4,a_5\in I(a)$ such that $x=a_4-a_0=a_0-a_5$. Thus $2a_0=a_4+a_5\in C$. It returns to the first question.

3. I think we should impose some conditions on $a$, such as $I(a)$ is a subgroup of $A$. To make the answer to the first two questions are yes, what other conditons can be considered?

Answer 1

Let $A=\Bbb Z$ and $f\colon A\to \{,,\}$ given by $f(7)=$, $f(3)=f(11)=f(100)=$, and $f(x)=$ for all other $x$. Then for $a_0:=7$, we have $I(a_0)=\{\,x\in\Bbb Z\mid f(x)=\,\}=\{a_0\}$ as desired. With $a:=3$, we have $I(a)=\{3,11,100\}$, hence $C=\{6,14,22,103,111,200\}$. We see that $2a_0\in C$, but $I(a)-a_0=\{-4,4,93\}$ differs from $a_0-I(a)=\ne\{-93,-4,4\}$ but they are not disjoint.

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Re the third question: There can only be one $a$ such that $0\in I(a)$, hence all that decoration with a function $f$ can be reduced to: Let $H$ be a subgroup of the abelian group $A$ and $a_0\in A$. If $2a_0=h_1+h_2$ with $h_1,h_2\in H$, then clearly $H=H+(h_1+h_2)=H+2a_0$ and hence $H-a_0=H+a_0=(-H)+a_0$.