If $G, H$ are approximate groups that $2G\cap 2H$ is also approximate group

Question

Let $G$ be a $K$-approximate group in an ambient group $Z$, and let $H$ be a $K'$-approximate group in $Z$. Show that $2G\cap 2H$ is a $(KK')^3$-approximate group.

Attempt of proof: First of all it is obvious that $0\in 2G\cap 2G$ and $(2G\cap 2H)=-(2G\cap 2H)$.

We need to show that $2(2G\cap 2H):=(2G\cap 2H)+(2G\cap 2H)\subseteq (2G\cap 2H)+X$ for some additive set $X$ such that $|X|\leq (KK')^3$.

1. If $x\in (2G\cap 2H)+(2G\cap 2H),$ then $x=x_1+x_2$ where $x_1,x_2\in 2G,2H$. Hence $x\in 4G\subseteq G+3X_1$ and $x\in 4H\subseteq H+3X_2$ where $|X_1|\leq K, |X_2|\leq K'$. We have shown that $$(2G\cap 2H)+(2G\cap 2H)\subseteq (G+X)\cap(H+Y)$$ for some additive sets $X,Y$ such that $|X|\leq K^3, |Y|\leq (K')^3$.

2. We notice that $$(G+X)\cap (H+Y)=\bigcup\limits_{x_0\in X, y_0\in Y} (G+x_0)\cap (H+y_0).$$ Fix some $x_0\in X, y_0\in Y$ and if $z\in (G+x_0)\cap (H+y_0),$ then $z=g+x_0=h+y_0$ for some $g\in G, h\in H$. So $g-h=y_0-x_0\in G-H$ and hence $y_0-x_0=g'-h'$ for some $g'\in G, h'\in H$. Then $g-h=g'-h' \Rightarrow$ $\beta:=g-g'=h-h'\in 2G\cap 2H$.

We have $z=\beta+x_0+g'\in (2G\cap 2H)+x_0+g'$ and we notice that $g'\in G$ depends on $x_0$ and $y_0$.

We have shown that $(G+x_0)\cap (H+y_0)\subseteq (2G\cap 2H)+x_0+g'$.

Question: Using the above results I need to show that $2(2G\cap 2H)\subseteq (2G\cap 2H)+X$ for some $X$ such that $|X|\leq (KK')^3$. Intuitively I know that this is true because $g'$ depends on $x_0, y_0$ but I do not know how to prove it rigorously.

I have tried to make this rigorous but I failed. I'd be thankful for any help! Thank you so much!

Answer 1

You're almost done - just note that the size of $\{x_0+g'(x_0,y_0) : x_0\in X, y_0\in Y\}$ trivially is at most $\lvert X\rvert\lvert Y\rvert$, since fixing a choice of $x_0,y_0$ fixes $x_0+g'(x_0,y_0)$.