TITLE: Monster coefficients
QUESTION [6 upvotes]: For three irreducible characters $\phi,\psi,\rho$ of a finite group $G$ define the Kronecker multiplicities as: 
$$g(\phi,\psi,\rho) = \langle \phi,\psi\cdot\rho\rangle
$$
where 
$$\langle \chi,\eta\rangle = \frac{1}{|G|}\sum_{x\in G} \chi(x)\,  \overline{\eta(x)}
$$
and $[\psi\cdot\rho] (x) = \psi(x) \rho(x)$ is the usual product. 
I am interested in Kronecker multiplicities for the Monster group $M$.  While the group is large, there are only 194 conjugacy classes.  
$$(1) \qquad \max_{\phi,\psi,\rho} g(\phi,\psi,\rho)$$
$$(2) \qquad \sum_{\phi,\psi,\rho} g(\phi,\psi,\rho)$$
$$(3) \qquad \sum_{\phi,\psi,\rho} g(\phi,\psi,\rho)^2$$
These sums are over all triples of irreducible characters, but because of the symmetries only about 1/6 of them need to be computed to get the answer.  If you can do this, I would also be curious about the specific characters maximizing (1).
The computation is beyond my computer algebra skills, but I know that GAP has the whole character table of $M$ ready to use.

REPLY [6 votes]: In GAP, you could simply iterate in a triple loop over $\phi,\psi,\rho$, calculate the $g$-values and find maximum and sum values:
ct:=CharacterTable("M");
irrs:=Irr(ct);

m:=0; s:=0;q:=0; # max, sum, sumsquare

for rho in irrs do
  for psi in irrs do
    ten:=rho*psi; # tensor product
    for phi in irrs do
      g:=ScalarProduct(phi,ten);
      if g>m then m:=g;fi;
      s:=s+g;
      q:=q+g^2;
    od;
  od;
od;

Afterwards look at the values of m, s, and q. Unless I have mistyped something, the results I get are (for the monster group):

Maximum is $21458051228477513179513856=2^{10}\cdot281\cdot443\cdot599\cdot6571\cdot42768299767$
Sum is $247017097351847432984363535932$ (Thank you, @James for the correction)
Sum-Squares is $808017424794512875894769468067441075690144312450960558$ (ditto corrected, also typo fixed)