TITLE: Nilpotent elements of group algebra $\Bbb CG$
QUESTION [6 upvotes]: Goal: explicitly find a nilpotent element of the group algebra $\Bbb C G$ for some finite group $G$. This exists if and only if $G$ is non abelian by Maschke's theorem and Wedderburn-Artin.

By Maschke's theorem, the group algebra $\Bbb C G$ is semisimple for finite $G$. So for any non-abelian group $G$, there is a nontrivial Wedderburn component, and thus  nilpotent elements.
Take $G=D_{8}$, the dihedral group or order $8$. Then $G$ is non-abelian and we know that the irreducible representations have dimensions $1,1,1,1$, and $2$. Thus
$$\Bbb C G\cong \Bbb C^{\oplus 4}\oplus M_2(\Bbb C)$$
and under this map we have nilpotent elements like
$$\left(0,0,\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} \right).$$
In theory, then, we should be able to find nilpotent elements in $\Bbb C G$. We know they exist, but we don't know the actual correspondence $\Bbb C G\tilde \to \Bbb C^{\oplus 4}\oplus M_2(\Bbb C)$.

My only lead: there is a well-known result we can use to calculate central primitive idempotents. For $D_8$, with the presentation
$$D_8=\langle a,b: a^4=b^2=abab=1\rangle,$$
the element corresponding to the $2\times 2$ identity matrix (i.e., the element $(0,0,\text{Id}_2)$), is
$$\frac{1}{2}(1-a^2).$$
But this is not really helpful, and I still have no leads on how to get the element $\left(0,0,\begin{pmatrix}0,& 1\\ 0 & 0\end{pmatrix}\right)$ for example.

I have done all of the guess work I can. For the case $G=D_8$, a nilpotent must square to $0$, so I checked elements of the form $\sum_{G}^8c_ig$ for $c_i\in\{-3,-2,1,0,1,2,3\}$ with no luck. If there are no "easy" nilpotent elements, then I will need to get more clever with the search.

Note: there is nothing special about $D_8$. I just picked it because the two dimensional representation was easy to start playing with by hand in the first place.

REPLY [3 votes]: Just to provide more context.
Let $G$ be a finite group, and let $H\le G$ be a non-trivial subgroup, which is not normal.  Then if 
$$ \alpha_H=\frac{1}{|H|}\sum h$$
you can check that $\alpha_H^2=\alpha_H$, and thus $(1-\alpha_H)\alpha_H=0$.  Let $g\in G$ be such that $g\notin N_G(H)$. Then we have 
$$ \alpha_Hg\neq\alpha_Hg\alpha_H $$
and so if we define $\beta_H=\alpha_Hg(1-\alpha_H)$, then $\beta_H\neq0$ but $\beta_H^2=0$.
This only works if we can find a non-normal subgroup $H$.  The finite groups that have every subgroup normal are well-known, and the non-abelian ones look like $A\times Q_8$, where $A$ is abelian and $Q_8$ is the quaternion group. So to finish this argument for all non-abelian groups, it suffices to exhibit a nilpotent element in $\mathbb{C}Q_8$.
I'll leave this to you, only noting that you cannot do it over $\mathbb{Q}$, since $\mathbb{Q}Q_8$ has no nontrivial nilpotents. But it can be done over $\mathbb{Q}(i)$.