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\begin{document}

\section{Equation 3308}

Equation 3308 is $x\diamond y=x\diamond(x\diamond (y\diamond x))$.  Setting $y=hx$ and $a\diamond b=f(ba^{-1})a$ for some map $f\colon G\to G$, we get the functional equation
\[
f^2(f(h^{-1})h)=f(h).
\]
Think of this as saying that once we know $(a,b),(a^{-1},c)\in f$, then there is some $d\in G$ such that $(ca,d),(d,b)\in f$.  Of course, by symmetry in the inverses, there is also some $e\in G$ such that $(ba^{-1},e),(e,c)\in f$.


Define $\mathscr{E}$ as the collection of sets $E\subseteq G^2$ satisfying the following properties.
\begin{itemize}
    \item[(1)] $E$ is finite.
    \item[(2)] $E$ is a function with $1\in {\rm dom}(E)$.
    \item[(3)] If $(a,b),(a^{-1},c)\in E$, then $(ca,d),(d,b),(ba^{-1},e),(e,c)\in E$ for some $d,e\in G$.
    \item[(4)] If $(a,b)\in E$ and $a^{-1}\notin {\rm dom}(E)$, then $ba^{-1},ab^{-1}\notin {\rm dom}(E)$.
    \item[(5)] If $(a,b),(c,d)\in E$, and $ab^{-1}=(cd^{-1})^{\pm 1}$, then $(a,b)=(c,d)$ or $(a,b)=(d,c)$.
    \item[(6)] If $(a,b),(b,a)\in E$, then $a^{-1}\in {\rm dom}(E)$ if and only if $b^{-1}\in {\rm dom}(E)$.
\end{itemize}

\begin{lemma}
For any $E\in \mathscr{E}$ and any $a\in G$, there is an extension $E\subseteq E'\in \mathscr{E}$ where the functional equation holds for $a$.
\end{lemma}
\begin{proof}
{\bf Case 1}: Assume $(a,b)\in E$ for some $b\in G$.

If $a^{-1}\in {\rm dom}(E)$, then by condition (3) we are already done.  So reduce to the case when $a^{-1}\notin {\rm dom}(E)$.  By condition (4), $ab^{-1},ba^{-1}\notin {\rm dom}(E)$.  By condition (2), $ab^{-1}\neq 1$.

There are two main subcases to consider.

{\bf Case 1a}: $(b,a)\in E$.

Note that $b^{-1}\notin {\rm dom}(E)$ by condition (6).  Take $c,d,e,f,g,h,i,j$ to be independent generators of $G$ (sorry for the re-use of $f$ and $h$) not appearing in the normal forms of the entries in $E$, and fix
\begin{eqnarray*}
E' & := & E\cup \{(a^{-1},c),(ca,d),(d,b),(ba^{-1},e),(e,c),(b^{-1},f),(fb,g),(g,a),(ab^{-1},h),(h,f),\\
 & & \quad \quad (hba^{-1},i),(i,e),(eab^{-1},j),(j,h)\}.
\end{eqnarray*}
Conditions (1) and (2) are clear.  Condition (3) is also satisfied, using the fact that the new letters are independent and that $ab^{-1}\neq 1$.

For condition (4), first note that if $(x,y)\in E$ with $x^{-1}\notin {\rm dom}(E)$, then $xy^{-1},yx^{-1}\notin {\rm dom}(E)$ since $E$ satisfies condition (4).  Thus, the only way that at least one of them is in ${\rm dom}(E')$ is if one of them equals $ab^{-1}$ or $ba^{-1}$.  By (5), then $(x,y)=(a,b)$ or $(x,y)=(b,a)$, which is okay.  For the new pairs in $E'$, condition (4) is checked directly (using $a,b,ab^{-1},ba^{-1}\neq 1$).

Condition (5) for $E'$ is quickly verified, by directly checking the five new pairs.

{\bf Case 1b}: $(b,a)\notin E$.

Now just take $c,d,e$ to be independent generators of $G$ not appearing in the normal forms of the entries in $E$, and fix
\begin{eqnarray*}
E' & := & E\cup \{(a^{-1},c),(ca,d),(d,b),(ba^{-1},e),(e,c)\}.
\end{eqnarray*}
All the conditions are verified essentially like before.

{\bf Case 2}: Assume $a\notin {\rm dom}(E)$.  If $a^{-1}\in {\rm dom}(E)$, then we can repeat Case 1 for $a^{-1}$, and are done (since it makes the functional equation hold for $a^{-1}$ and its inverse).

So, assume $a,a^{-1}\notin{\rm dom}(E)$.  If there is a pair $(x,y)\in E$ with $xy^{-1}=a^{\pm 1}$, then apply Case 1 to $x$, and we reduce to a previous work.  So, we may as well assume there is not such pair $(x,y)$.  Then by passing to $E\cup\{(a,b)\}$, with $b$ a generator of $G$ not appearing anywhere in $E$ or $a$, this still satisfies all the conditions and we again reduce to a previous case (and note that we don't break condition (4) because of the previous sentence).
\end{proof}

The functional equation for 3456 is $f(1)=f(f(f(1)^{-1})f(1))$. Using the seed
\begin{eqnarray*}
\{(1,x_1),(x_1,x_2),(x_2,x_1),(x_1^{-1},x_3),(x_3x_1,x_4),(x_4,x_2),(x_2x_1^{-1},x_5),(x_5,x_3),(x_2^{-1},x_6),\\ \ (x_6x_2,x_7),(x_7,x_1),(x_1x_2^{-1},x_8),(x_8,x_6),(x_8x_2x_1^{-1},x_9),(x_9,x_5),(x_5x_1x_2^{-1},x_{10}),(x_{10},x_8)\}
\end{eqnarray*}
and extending to a full function using the previous lemma, we are done.


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