Suppose $G$ is a group, $S \subset G$. Let’s call $S$ sum-free iff $\forall a, b \in S$ we have $ab \notin S$. What is the value of $c := \sup\{\frac{|A|}{|G|}| $G$ \text{ is a finite group, } A \subset G \text{ is sum-free}\}$?
It is not hard to see, that $c \geq \frac{1}{2}$. Indeed, in order to achieve it one can simply take any group $G$ with a subgroup $H$ of index $2$. $G \setminus H$ will be sum-free in that case. However, I suspect that $c$ can be much larger…
A sum-free subset $S$ of $G$ cannot have more than $\frac{1}{2}|G|$ elements. Thus $c\leq\frac{1}{2}$.
Proof. Indeed, fix $s\in S$ and consider
$$f:G\to G$$ $$f(g)=gs$$
Since $S$ is sum-free then $f(S)\subseteq G-S$ which obviously implies $|f(S)|\leq |G-S|$. Since $f$ is a bijection then $|S|=|f(S)|$ and thus $$|S|=|f(S)|\leq |G-S|=|G|-|S|$$ which finally leads to $$|S|\leq\frac{1}{2}|G|$$