It is easy to see that the chromatic number $\chi(G)$ of graph $G$ of order $n$ and the independence number $\alpha(G)$ are related by the inequality $\alpha(G)\geq n/\chi(G)$.
Therefore if $\chi(G)=3$, the graph $G$ has an independent set of size $\geq n/3$.
The alternating group graph $AG_n$ is a Cayley graph of the alternating group $A_n$ given by $S=\{(1,2,k)^{\pm1}| k=3,\ldots,n\}$.
Thus, $AG_n=\operatorname{Cay}(A_n,S)$ is
the graph with vertex set $A_n$ and edge set $\{(\tau,\tau\sigma)\,\mid\,\tau\in A_n,\sigma\in S\}$.
Hence if $\chi(AG_n)=3$, then $\alpha(AG_n)\geq n!/6$.
However, we know that $\alpha(AG_3)=1$, $\alpha(AG_4)=4$, $\alpha(AG_5)=20$, and $\alpha(AG_6)=120$,
but the value for $\alpha(AG_7)$ is apparently not known (Stan Wagon, pers. comm., June 27, 2022), (see also Alternating group graph).
Addendum.
To complete the picture, some simple remarks about the coloring of the graph $AG_n$ for $n=4,5,6$.
$n=4$. Let $V$ be a subgroup of Klein in $A_4$. The sets $V$, $V(123)$, and $V(132)$ are independent since $V$ does not contain cycles of length $3$.
$n=5$. Let $D=\operatorname{gr}((12345),(15)(24)$.
The subgroup $D$ has order $10$ and obviously contains no cycles of length $3$.
Also let $S=\{(12k)^{\pm1}\mid k=3,4,5\}$.
- Choose three elements $v,x,y$ in the group $A_5$ with the following properties:
$(i)$ $Dv\cap S=\varnothing$;
$(ii)$ $x^{-1}Tx\cap S$=$y^{-1}Ty\cap S=\varnothing$, where $T$ is a set cycles of length $3$ in $Dv$.
$(iii)$ $D,Dv,Dx,Dvx,Dy,Dvy$ is a complete set of right cosets of $D$ in $A_n$.
This requires a little computation. I have consistently found that one can take
$v=(13)(25)$, $x=(12)(34)$, $y=(12)(45)$.
Note that following the GAP package we multiply permutations from left to right, that is, $(12)(13)=(123)$.
Since $D$ contains no cycles of length $3$, the set $Da$ is independent for any permutation $a\in A_5$.
The sets $V_1=D\cup Dv$, $V_2=Dx\cup Dvx$, and $V_3=Dy\cup Dvy$ are independent.
Prove, for example, that $V_2$ is independent.
If $(u,w)$ is an edge connecting vertices of $V_2$, then we can assume
$u\in Dx$, $w\in Dvx$ and $u^{-1}w\in S$.
If $u=ax$, $w=bvx$, then $u^{-1}w=x^{-1}a^{-1}bvx\in S$.
Therefore $a^{-1}bv\in Dv$ is a cycle of length $3$ and we get a contradiction with 1$(i)$.
We have $V_i$ are independent, $|V_i|=20$, and $A_5=\cup_i V_i$.
$n=6$.
Let $H=\operatorname{gr}((12)(34), (134)(256)$.
The subgroup $H$ is isomorphic to the group $A_5$ and contains no cycles of length $3$.
Also let $S=\{(12k)^{\pm1}\mid k=3,4,5,6\}$.
It turns out that we can choose three permutations satisfying the properties 1$(i,ii,iii)$:
$v=(35)(46)$,
$x=(12)(35)$,
$y=(12)(36)$.
Then, by the same scheme as above, we can prove that
The sets $V_1=H\cup Hv$, $V_2=Hx\cup Hvx$, and $V_3=Hy\cup Hvy$ are independent
and hence we can color our graph $AG_6$ in three colors.
And of course $|V_i|=120$.
Addendum 2.
$n=7$.
The chromatic number of the graph $AG_7$ is $4$.
Since we know, thanks to Gordon Royle, that $\chi(AG_7)>3$
we only need to establish that this graph can be properly colored in $4$ colors.
My computation was based on similar considerations as for the graphs $AG_5$ and $AG_6$.
In the group $A_7$ there is a subgroup of order $168$,
which does not contain any cycles of length $3$.
Let us denote this group by $H$. We can check that
$H=\operatorname{gr}((1,4)(2,3), (2,4,6)(3,5,7) )$.
Therefore each right coset $Hx$, $x\in A_7$ is an independent set.
Hence, in turn, it follows that to check the independence of the set $V=Hx\cup Hy\cup\ldots$
it is sufficient to check that $H\cap xSy^{-1}=\varnothing$
for each pair of coset representatives lying in $V$, where
$S=\{(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,1,3),(2,1,4),(2,1,5),(2,1,6)\}$.
I have found a collection of $4$ independent disjoint sets in $A_7$,
each of them is a union of several right cosets:
\begin{eqnarray*}
V_1 &=& H+H(4,5,6)+H(4,6)(5,7)+H(3,4,6,7,5)+H(3,5,4,6,7),\\
V_2 &=& H(5,6,7)+H(4,5)(6,7)+H(4,7,6)+H(3,4,7,6,5),\\
V_3 &=& H(5,7,6)+H(4,6,5)+H(4,7)(5,6),\\
V_4 &=& H(4,5,7)+H(4,6,7)+H(3,4,5).
\end{eqnarray*}
(Here I use $'+'$ instead of $'\cup'$.)
Already after I found these sets I realized that they could be computed
by a complete enumeration of all subsets of the 15 right cosets
of the subgroup $H$ in group $A_7$.
The total number of such subsets is a little more than 30 thousand.
Here are commands from the GAP package,
which allow you to check the independence of each of $V_i$ and
that they do not overlap.
And the cardinality of these sets are respectively: 840, 672, 504, 504.
- We compute the order of the subgroup $H$ and the fact that $H$ does not contain cycles of length $3$:
G:=AlternatingGroup(7);;
H:=Subgroup(G,[(1,4)(2,3), (2,4,6)(3,5,7)]);;
Order(H);
Filtered(H,x->Order(x)=3 and Order(Centralizer( G, x ))>9);
- Check that $V_i$ are independent sets.
Let $R_i$ be the representatives of the right cosets lying in $V_i$
(here for $V_1$, for $V_i$ we should replace
Combinations(R1,2) by Combinations(R2,2) and so on):
R1:= [ (), (4,5,6), (4,6)(5,7), (3,4,6,7,5), (3,5,4,6,7)];;
R2:= [ (5,6,7), (4,5)(6,7), (4,7,6), (3,4,7,6,5) ];;
R3:= [ (5,7,6), (4,6,5), (4,7)(5,6) ];;
R4:= [ (4,5,7), (4,6,7), (3,4,5) ];;
S:=[(1,2,3),(1,2,4),(1,2,5),(1,2,6),(1,2,7),(2,1,3),(2,1,4),(2,1,5),(2,1,6),(2,1,7)];;
for a in Combinations(R1,2) do
x:=a[1];
y:=a[2];
q:=IsEmpty(Intersection2(H, x*S*y^-1));
if not q then break; fi;
od;
q;
- We compute the cardinality of sets.
It suffices to check that all cosets are pairwise distinct.
At the same time we check that the sets do not overlap:
R:=Concatenation(R1,R2,R3,R4);;
V:=List(R,x->H*x);;
V:=AsSSortedList(V);;
Size(V);
Note.
Apparently for the computation of independent sets the graph whose vertices are right cosets of $H$ in $A_7$ and two vertices are connected by an edge can be useful if there is at least one edge between the elements of these cosets in the graph $AG_7$. I will try to draw this graph and post it here.
Addendum 3. (01.07.2022)
And this is the above graph with its $4$-vertice colouring.
