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Consider $G$ be an abelian group with $A$ and $B$ are subsets of $G$. Consider $A + B = \{a + b: a\in A, b \in B\},\quad A - B = \{a - b: a\in A, b\in B\}$

If $A = B$, we have A + A = \{a + a': a, a'\in A\} and A - A = \{a - a': a, a'\in A\}. For all $a, b$ distinct elements of $G$, we get $a - b$ is not equal to $a - b$ unless $a - b$ has order $2$.

Now the question is how to construct sets with more additions then subtractions (differences) for the finite set $A$ with $|A + A| > |A - A|$, where $|A|$ is a cardinality set of $A$.

Also, I want the proof of second part, For any set of integers $A$, for every integer $x$, how we define $x \cdot A = \{xa: a \in A\}$. If $A$ is set and can be expressible in more additions then subtractions for $x$ not equal to zero and for any integer $y$, the set becomes $x \cdot A + \{y\} = \{xa + y: a \in A\}$ is also an more additions than subtractions.

my final question is: How to prove the first explicit constructions of infinite families of (more additions than subtractions) sets of integers by using the probability method to prove the existence of such sets in certain finite abelian groups. in certain finite abelian groups?

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    yes! I mean addition and subtraction is nothing but sums and differences. I got it where I am wrong...2011-08-15

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If you type "more sums than differences" into Google, you will be directed to several papers on this topic.

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    well! I will see as per your instruction.2011-08-15