# Discrete Mathematics/SETS

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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Please try to format your writing so that it is possible to read. Be sure to make proper line breaks and preferably use $\LaTeX$ markup for the math parts. Furthermore $a-b$ is always $a-b$. In every abelian group, for every $a,b$. I'd guess there is a typo there, but I did not want to alter the original text other than formatting and $\LaTeX$. –  Asaf Karagila Aug 15 '11 at 9:16
A remark to the first question. This is not possible, if $G$ is elementary 2-abelian. In that case addition and subtraction mean the same thing, so the inequality $|A+A|>|A-A|$ is always false. IOW is the group $G$ given to us, or are we allowed to vary it? –  Jyrki Lahtonen Aug 15 '11 at 9:18
yes! I mean addition and subtraction is nothing but sums and differences. I got it where I am wrong... –  Gandhi Aug 15 '11 at 12:29