TITLE: Inverse of $1-g-g^{-1}$ in group ring $\mathbb{Z}[G]$ for $o(g)=5$.
QUESTION [6 upvotes]: I'm trying to prove that if $G$ is a group and $g\in G$ is an element of order $5$ then $x=1-g-g^{-1}$ is an unit element in the group ring $\mathbb{Z}[G]$. I'm trying to find an explicit inverse  by computing $xg^r$ for $r=1,2,3,4$ and trying to add them in a good way to get $1$. But so far I can't find a good "combination". Any hints on how to compute an inverse for this element?.
[EDIT]
I found a solution for this but it is probably the hardest way. I'm trying to find an inverse of the form $a+bg+cg^2+dg^3+eg^4$, so i take the equation $x(a+bg+cg^2+dg^3+eg^4)=1$ and it becomes a linear algebra problem. I found by solving the $5\times 5$ integer linear system that $a=1,b=-1,c=0,d=-1,e=0$ is a solution, i.e., $x(1-g-g^4)=1$. 
I think there should be other ways to approach these kind of exercises.

REPLY [3 votes]: We have $x=1-g-g^4$. The extended Euclidean algorithm for $\gcd(1-g-g^4,g^5-1)$ gives
$$
1 = (1 - g^2 - g^3)(1-g-g^4)+(-g - g^2)(g^5-1)
$$
Therefore, the inverse of $x$ is $1 - g^2 - g^3$ because $g^5-1=0$.