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.


You can get a lot more insight by actually looking at what the 2-dimensional representation is explicitly. It's just the usual action of $D_8$ as symmetries of a square, so $a$ maps to the matrix $A=\begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix}$ and $b$ maps to $B=\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$. We can now just fiddle about to find a combination of these matrices that gives $\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$. Specifically, we have $$BA=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$$ so $$\frac{BA-A}{2}=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}.$$

Now this doesn't tell us that $\frac{ba-a}{2}$ is nilpotent, only that its image in the 2-dimensional representation is nilpotent. We can eliminate its image in the other representations by multiplying by the idempotent $\frac{1-a^2}{2}$ which you found. So, we conclude that $$\frac{ba-a}{2}\cdot\frac{1-a^2}{2}$$ is a nontrivial nilpotent element of $\mathbb{C}D_8$.

Just to verify, we can square it: $$\begin{align*} \left(\frac{ba-a}{2}\cdot\frac{1-a^2}{2}\right)^2 &= \left(\frac{ba-a}{2}\right)^2\cdot\left(\frac{1-a^2}{2}\right)^2 \\ &=\frac{baba-ba^2-aba+a^2}{4}\cdot \frac{1-a^2}{2} \\ &=\frac{1-ba^2-b+a^2}{4}\cdot \frac{1-a^2}{2} \\ &=\frac{1-a^2-ba^2+ba^4-b+ba^2+a^2-a^4}{8} \\ &=\frac{1-a^2-ba^2+b-b+ba^2+a^2-1}{8} \\ &= 0. \end{align*} $$

  • 1
    $\begingroup$ Thank you this is exactly what I was looking for! My and my roommate have been going insane because we really thought this should be easier. I originally thought that the element $ba-a$ would be nilpotent because of that exact logic, and then started questioning everything when I saw it wasn't. Makes sense now. $\endgroup$ – Elliot G Sep 2 '18 at 21:43

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 $\mathbb{Q}(i)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.