I'm reading some lecture notes and get stuck on one detail. We wish to prove the following:
(1) Let $\alpha > 0$ and $A \subseteq [N]$ be of size $\geq$ $\alpha N$. Then $A + A + A$ contains an arithmetic progression $P$ of size $$ \vert P \vert \geq c\alpha N^{c \alpha^3} $$
Here, $c > 0$ is some absolute constant. Now, instead of proving the above we prove the following:
(2) Let $\alpha > 0$ and $A \subset \mathbb{Z}_M$ ,for some prime $M$, be of size $\alpha M$. Then $A + A + A$ contains an arithmetic progression $P$ of size $$ \vert P \vert \geq \frac{1}{2\pi}\alpha M^{\alpha^3/4} $$
The notes now says that (1) easily follows from (2) (we should only need $M$ to be a prime between $6N$ and $12N$.) However I don't manage to see how. Any hints on how to show that (2) implies (1)? I was thinking about looking at $P \cap N$, with $P$ as in (2), but did not get anywhere. And where does $M$ being prime enter the picture?
The point is that the arithmetic progression in the cyclic group is also an arithmetic progression in the integers.
Lemma: Let $B \subset [L]$ and let $p > 2L$. If $(b_0, \dots, b_k)$ with $b_i\in B$ is an arithmetic progression in $\mathbb{Z}_p$, then $(b_0, \dots , b_k)$ is an arithmetic progression in $\mathbb{Z}$.
Proof: Suppose there is some $0 \le d \le p-1$ such that $b_i = b_{i-1} + d \pmod{p} $ for each $i$, that is it is a progression in $\mathbb{Z}_p$. Without loss we can assume $1\le d \le (p-1)/2$, else replace $d$ by $p-d$ and reverse the order of the progression.
Yet then $0 \le b_{i-1}+ d \le L + (p-1)/2 < p$, thus $b_{i-1} + d = b_i$ as integers. $\square$
Now only note that $A+A+A \subset [3N]$ and $M > 2 (3N)$. The upper bound on $M$ ensures that you can just replace the $M$ by $N$ in the bound (adjusting the constants).