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\begin{proof}
  Choose $X \subseteq \mdiff 2A - {}A$ maximal
  such that the translates $x \plus A$ with $x \in X$
  are disjoint. Such a set $X$ cannot be too
  large: $\forall x  \in X$, $x \plus A \subseteq
  \mdiff 3A - {}A$, so by \nameref{plunn}, since
  $|\mdiff 3A - {}A| \le K^4 |A|$,
  \[
    |X| |A| = \left| \bigcup_{x \in X} (x + A) \right|
    \le |\mdiff 3A - {}A|
  .\]
  So $|X| \le K^4$. We next show
  \[
    \mdiff 2A - {}A \subseteq X \plus A \minus A
    \tag{$*$}
    \label{fr_wts_eq}
  .\]
  Indeed, if $y \in \mdiff 2A - {}A$ and $y \notin
  X$, then by maximality of $X$, $y \plus A \cap x
  \plus A
  \neq \emptyset$ for some $x \in X$ (and if $y
  \in X$, then clearly $y \in X \plus A \minus
  A$).

  It follows from \eqref{fr_wts_eq} by induction that
  $\forall l \ge 2$,
  \[
    \mdiff lA - {}A \subseteq (l - 1)X + A - A
    \tag{$**$}
    \label{fr_la_a}
  ,\]
  since
  \[
    \mdiff lA - {}A = A \plus \ub{(l - 1)A \minus
    A}_{\subseteq (l - 2)X \plus A \minus A} \subseteq (l -
    2) X \plus \ub{2A - A}{\subseteq X \plus A \minus
    A} \subseteq (l - 1)X \plus A \minus A
  .\]
  Now let $H \le \Fp^n$ be the subgroup generated
  by $A$, which we can write as
  \[
    H = \bigcup_{l \ge 1} (\mdiff lA - {}A)
    \stackrel{\eqref{fr_la_a}}{\subseteq} Y + A - A
  \]
  where $Y \le \Fp^n$ is the subgroup generated by
  $X$.

  But every element of $Y$ can be written as a sum
  of $|X|$ elements of $X$ with coefficients
  amongst $0, 1, \ldots, p - 1$, hence $|Y| \le
  p^{|X|} \le p^{K^4}$. To conclude, note that
  \[
    |U| \le |Y| |A \minus A| \le p^{K^4} \le
    p^{K^4} K^2 |A|
  ,\]
  where we use \nameref{plunn} or even
  \nameref{ruzsa_ineq}.
\end{proof}

\begin{example}
  % Example 1.12
  Let $A = V \cup R$ where $V \le \Fp^n$ is a
  subspace of dimension $K \ll d \ll n - K$ and
  $R$ consists of $K - 1$ linearly independent
  vectors not in $V$.

  Then
  \[
    |A| = |V \cup R| = |V| + |R|
    = p^{n / k} + K - 1 \sim p^{n / k} = |V|
  \]
  and
  \[
    |A + A| = |(V \cup R) + (V \cup R)|
    = |V \cup (V + R) \cup (R + R)| \sim K|V|
  .\]
  But any subspace $K \le \Fp^n$ containing $A$
  must have size at least $p^{n / K + (K - 1)}
  \sim |V| \cdot p^K$, so the exponential
  dependence on $K$ is necessary.
\end{example}

\begin{fcthm}[Polynomial Freiman-Ruzsa, due to Gowers--Green--Manners--Tao 2024]
  % Theorem 1.12
  Assuming:
  - $A \subseteq \Fp^n$
  - $|A + A| \le K|A|$
  Then:
  there exists a subspace $K \le \Fp^n$ of size at
  most $C_1(K) |A|$ such that for some $x \in
  \Fp^n$,
  \[
    |A \cap (x + K)| \ge \frac{|A|}{C_2(K)}
  ,\]
  where $C_1(K)$ and $C_2(K)$ are polynomial in
  $K$.
\end{fcthm}

\begin{proof}
  Omitted, because the techniques are not relevant
  to other parts of the course. See Entropy
  Methods in Combinatorics next term.
\end{proof}

\begin{fcdefn}[]
  % Definition 1.13
  \glssymboldefn{energy}%
  \glsnoundefn{adde}{additive energy}{additive
  energies}%
  \glsnoundefn{addq}{additive quadruple}{additive
  quadruples}%
  Given $A, B \subseteq G$ we define the
  \emph{additive energy} between $A$ and $B$ to be
  \[
    E(A, B) = |\{(a, a', b, b') \in A \times A
    \times B \times B : a + b = a' + b'\}|
  .\]
  We refer to the quadruples $(a, a', b, b')$ such
  that $a + b = a' + b'$ as \emph{additive
  quadruples}.
\end{fcdefn}

\begin{example}
  % Example 1.14
  Let $V \le \Fp^n$ be a subspace. Then
  $\energy(V) = \energy(V, V) = |V|^3$.

  On the other hand, if $A \subseteq \Zp$ is
  chosen at random from $\Zp$ (each element chosen
  independently with probability
  $\alpha > 0$), then with high probability
  \[
    \energy(A) = \energy(A, A) = \alpha^4 p^3 =
    \alpha |A|^3
  .\]
\end{example}

\begin{fclemma}[]
  \label{lemma_1_16}
  % Lemma 1.16
  Assuming:
  - $A, B \subseteq G$
  - both non-empty
  Then:
  \[
    \energy(A, B) \ge \frac{|A|^2 |B|^2}{|A + B|}
  .\]
\end{fclemma}

\begin{proof}
  \glssymboldefn{r}%
  Define $r_{A + B}(x) = |\{(a, b) \in A \times B
  : a + b = x\}|$ (and notice that this is
  the same as $|A \cap (x \minus B)|$). 
  Observe that
  \begin{align*}
    \energy(A, B)
    &= |\{(a, a', b, b') \in A^2
    \times B^2 : a + b = a' + b'\} \\
    &= \sum_{x \in G} r_{A \plus B}(x)^2 \\
    &= \sum_{x \in A \plus B} r_{A \plus B}(x)^2 \\
    &\ge \frac{\left( \sum_{x \in A \plus B} r_{A
    \plus
    B}(x) \right)^2}{|A \plus B|}
    &&\text{by Cauchy-Schwarz}
  \end{align*}
  but
  \begin{align*}
    \sum_{x \in G} |A \cup (x \minus B)|
    &= \sum_{x \in G} \sum_{y \in G}
    \indicator{A}(y) \indicator{x \minus B}(y) \\
    &= \sum_{x \in G} \sum_{y \in G}
    \indicator{A}(y) \indicator{B}(x - y) \\
    &= |A||B|
  \end{align*}
  (As usual, $\indicator{A}$ here means the indicator
  function).
\end{proof}