# number of sum free subsets

Given an abelian group $$G$$ of order $$n$$, how I do show that the number of sum free subsets of G is at most $$2^{(n/2 + o(n))}$$. Sum free subsets meaning $$A\subseteq G$$ s.t. $$\forall x,y,z \in A, x + y \neq z$$ I believe there is a proof by Alon but I do not understand it.
1.) Given the Cayley Graph on group G with respect to S $$H(G,S)$$, I do not see how it is a regular graph and
2.) He says $$\sum_{i=1}^{\log(n)} \binom{n}{i} = 2^{o(n)}$$, and the number of sum free subsets of cardinality greater than log(n) by theorem 1.2 is at most $$2^{(1+o(\log(n)^{-0.1})n/2}$$ so is $$2^{(1+o(\log(n)^{-0.1})n/2} + 2^{o(n)} \equiv 2^{(1+o(1))n/2}$$ ?
• If you want proof-explanation for Alon's approach, you should include it in the post and point out which part got you stuck. – Lee David Chung Lin Apr 21 at 4:23