This is an extension to my earlier question.
Is there a purely algebraic algorithm to find inverses in the group algebra? For example, in the group algebra $\mathbb{C}S_{4}$, how would one go about finding an inverse for $(1 2 3) + 2(1 2)(3 4)$.
By a purely algebraic algorithm, I mean one that avoids, if possible, the use of representation theory. (So if possible, it shouldn't require that I know the irreps of the group, and it shouldn't require that I switch to matrices.)
The answer to my earlier question pointed out that $(1 2 3) + 2(1 2)(3 4) = (1 2 3) (1 + 2(1 4 3))$ so it is sufficient to find an inverse for $1 + 2(1 4 3)$
Now, after a little experimentation, I've worked out how to do examples like this by inspection. The inverse is 1/9 of $1 - 2(1 4 3) + 4(1 3 4)$.
This is by analogy with the fact that the inverse of $1+x$ is $1-x+x^{2}-...$. Given $g$ of order $n$, then $(1+ag)(1-ag+a^{2}g^{2}-...\pm a^{n-1}g^{n-1}) = 1\pm a^{n}g^{n} = 1\pm a^{n}$, which is a scalar.
(Of course, some elements of the group algebra are zero divisors, and hence don't have inverses. For example $(1+(1 2)) (1-(1 2)) = 0$.)
Okay, so to get to the point. I'm stuck on how to proceed for elements of the form $1+ag+bh$, for example $1+2(1 2)+3(3 4)$. It's not clear that $ag+bh$ will be of finite order, so the above trick won't work.
Is there some analogue of Moebius inversion or principle of inclusion-exclusion that might be used?
EDIT: I think I have figured out how to do this (though I'm still curious whether there are other methods). It's easiest to explain with an example.
Suppose we want to invert $1+(1 2)+(3 4)$. It's clear that the answer, if it exists, will be of the form $a+b(1 2)+c(3 4)+d(1 2)(3 4)$. Multiplying these two expressions together, we get $a+a(1 2)+a(3 4)+b(1 2)+b+b(1 2)(3 4)+c(3 4)+c(1 2)(3 4)+c+d(1 2)(3 4)+d(3 4)+d(1 2)$. Now, identify coefficients with $1+0(1 2)+0(3 4)+0(1 2)(3 4)$, and we get a linear system in a, b, c, d. Solving the linear system gives us our inverse: 1/3 of $1+(1 2)+(3 4)-2(1 2)(3 4)$