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Let $G$ be some group, and let $\Bbb{C}G$ denote the complex group ring over $G$. Let $x,y \in \Bbb{C}G$, and define $[x,y] := xy-yx$ to be a (ring) commutator in $\Bbb{C}G$. Let $K$ be collection of all sums of commutators. I am reading a paper in which the authors claim that any element in $K$ takes the form

$$\displaystyle \sum_{j,k} a_{jk} g_k^{-1} w_j g_k$$

for some $w_j , g_k \in G$ and $a_{jk} \in \Bbb{C}$ with $\sum_{k} a_{jk}=0$ for every $j$.

I tried proving this several times, but I always end up with some truly dreadful (triple) sums. See page 9 of this paper if you need more information, or are curious how this little question connects to operator algebras. How does one go about proving the claim?

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Consider a "pure" commutator $[g,h]=gh-hg$ for $g,h\in G$. Then you can write $$gh-hg=h^{-1}(hg)h - g^{-1}(gh)g$$ which is of the form of your claim. If now you have to deal with a sum of pure commutators $$\sum_{j,k}a_{jk}[g_k,h_j]$$ then you repeat the procedure above for every pure commutator $$\sum_{j,k}a_{jk}[g_k,h_j] = \sum_{j,k}a_{jk}\left(h_j^{-1}(h_jg_k)h_j - g_k^{-1}(g_kh_j)g_k\right)$$ and the resulting sum of coefficients is always $0$ because every coefficient appears once with positive sign and once with negative. Finally, if you have a sum of arbitrary commutators, $$\sum_{j,k}a_{j,k}[x_k,y_j]$$ for $x_k,y_j\in\mathbb{C}G$ then you observe that any such commutator can be written as sum of pure commutators by bilinearity of the commutator: $$[x,y] = \left[\sum_{i} x_ig_i, \sum_{l}y_lh_l\right] = \sum_{i,l}x_iy_l[g_i,h_l],$$ and you are in the same situation as before.

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