.               

@C_R That helped, thank You. Any idea why the search engine would be targeted for login credentials?

You should inspect eq1,eq2,eq3. The fourth equation is redundant:

normal(eq1+eq2+eq3);

Your RootOf is in poor form and it is reducible. Reducible RootOfs are a no-no and shoud be subcased. Mixing RootOfs and radicals is no-no as well, thought I am unceratin of this one. You have much work to do here.

@acer ...odd. Somehow I ended up on EigenConditionNumbers page.

Thanks.

Are you going to post the problem itself?

@dharr Which Maple version is needed for this?

@sursumCorda Looking at the code, only linalg[iszero] uses Normalizer. So I must agree they've abandoned it.

@sursumCorda Why would you say identifying zero matrix is mathematically irrelevant?

@max125 I put in minor changes. It should be easier to read.

@Joe Riel A minor improvement:

map(proc(P) lprint(cat(whattype(eval(P)) , " " , P)); end, (sort([exports(GraphTheory, instance)]))): 

@dharr Infact there are 1455 subgroups and 56 conjugacy classes. I am hoping there is a way to generate them manually.

@sursumCorda Is this result correct?

F := proc(A) local a := A; while hastype(a, function) do a := subsindets(a, function, proc(F) [[op(0, F)], [op(F)]] end); end do; end;

F(cS(cK(cS(cI)))(cS(cK(cK))(cI))(x)(y));

[F(cS(x::anything)(y::anything)(z::anything) = x(z)(y(z))),
F(cK(x::anything)(y::anything) = x, cI(x::anything) = x)],
F(cS(cK(cS(cI)))(cS(cK(cK))(cI))(x)(y));
applyrule(%);
eval(applyrule([[f :: list, g :: list] = Fn(f, g)], %), Fn = ((L1, L2) -> op(L1)(op(L2))));

@NeraSnow This is the inverse of ifactors:
f := proc (L) local p; L[1]*mul(p[1]^p[2], p = L[2]) end proc;

So with substitutions it would be:
fsubbed := proc (L) local p; L[1]*mul(subs(args[2..-1], p[1])^p[2], p = L[2]) end proc;

example:
ifactors(10!);
f(ifactors(10!));
fsubbed(ifactors(10!), 2 = two, 3 = three);

@mmcdara There is an error in your definition of eq2. Compare to OP.