06:28
@MatheinBoulomenos so you suggest that that is not the kernels description?
I don't technically need a nice decsription, but it may come in handy later on
Let $x \in G^n$ for a finite non-abelian group, $x = (x_1, \dots, x_n)$. Can we somehow create a hom from $G^n$ into $\Bbb{Z}[G]$ that "has to do with" sums such as $x_1 + x_1 x_2 + \dots + x_1 \cdots x_n$?
Call such a sum $f(x), f : G^n \to \Bbb{Z}[G]$. It's definitely not a hom, but can we turn it into one say by doing $g(x) = \dfrac{1}{|G|} \sum_{g \in G} g f(x) g^{-1} - \text{ something} $ ?
If we do that exact expression, then $g(1) = 1$ so "Something" should be $1$ there.
It reminds me of fourier transform for finite groups, except I'm not working with complex numbers. I'm trying to keep this in $\Bbb{Z}[G]$ context
I mean $\dfrac{1}{n|G|}$ in that sum
I do know that when $x_1 + \dots + x_1 \cdots x_n = \sum_{g \in G} g$ then we have a hamiltonian cycle in the Cayley graph. That's why this is interesting.
So this hom $g$ would map into the kernel talked about above, and the homology at $\Bbb{Z}[G]$ module would measure something about Hamiltonian cycle problem in Cayley graphs, hopefully! :D
It's all speculation for now, but I am liking the looks of this symmetrizing sum thing
Note that $f$ preserves conjugation of its argument, so equivalently the sum becomes with $f(g x g^{-1})$
I did notice that conjugation doesn't view all the symmetries of $f(x)$. For instance if $G \simeq \Bbb{Z}_3, G = \{1, a, b\}, ab = 1$ then $f(x) = f(a,b) = a + ab$ but conjuate all you want and you don't necc get $b + ab$ in general (though you might here). So I should also through in the cycles of $(a,b,c, \dots, z) = x$
If you have a Hamilton cycle presented by $(x_1, \dots, x_n)$ then all cycle perms of that are also a Hamilton cycle presentation.
This is interesting because I haven't seen the problem broken down algebraically like this before, even though it's the most natural way to work at it
$\psi : G^n \to \Bbb{Z}[G], \ \ \psi(x) = \sum_{\sigma \in C_n} \sum_{g \in G} g f(\sigma x) g^{-1}$ maybe
Divided by $n|G|$ of course