I am looking for a group-algebra with a non-trivial solution to $x^2=x$. That is to say, a solution with $x\neq 1$ and $x \neq 0$ where $1$ is the identity.

We have $x\subset \mathbb{C}[G]$ for some group $G$ and want to find some small group such that $x^2=x$ has non-trivial solutions.

Is this possible and if so can you give an example of such a group? A simple group if possible.

  • $\begingroup$ Try some examples yourself! Literally any nontrivial finite group works. $\endgroup$ – Eric Wofsey Jun 16 at 3:43
  • $\begingroup$ What do you mean by simple group? en.m.wikipedia.org/wiki/Simple_group $\endgroup$ – Azlif Jun 16 at 3:44
  • $\begingroup$ @Eric. Hmm I got as far as complex numbers and the quaternion group and couldn't find anything. Maybe I missed something. $\endgroup$ – zooby Jun 16 at 3:44
  • $\begingroup$ @Azlif the usual definition. $\endgroup$ – zooby Jun 16 at 3:44
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    $\begingroup$ Hmm, maybe this question was too easy after all! Yes, I think even the group algebra $\mathbb{Z}_3$ will work!. $\endgroup$ – zooby Jun 16 at 3:49

Let $G=\{1,g\}$ be cyclic of order $2$. Then $x=\frac{1+g}{2}$ satisfies $$x^2=\frac{1}{4}(1+2g+g^2)=\frac{1}{4}(1+2g+1)=\frac{1+g}{2}=x.$$

Much more generally, it follows from the representation theory of finite groups that for any finite group $G$, $\mathbb{C}[G]$ is isomorphic to a product of matrix rings $M_n(\mathbb{C})$ for various values of $n$, with the number of factors in the product being the number of conjugacy classes in $G$. It follows immediately that if $G$ is any nontrivial finite group then there are nontrivial solutions to $x^2=x$ in $\mathbb{C}[G]$, since you can take an element that is $1$ on some of the factors and $0$ on others.

  • $\begingroup$ What do you mean by non-trivial? The quaternion group is non-trivial so I would expect there to be non-trivial solution in quaternion algebra. Maybe I missed it. $\endgroup$ – zooby Jun 16 at 3:54
  • $\begingroup$ Yes, there are plenty of solutions in the group algebra of the quaternion group. Note that the group algebra of the quaternion group is not the same thing as the usual algebra $\mathbb{H}$ of quaternions. $\endgroup$ – Eric Wofsey Jun 16 at 3:55
  • $\begingroup$ Hmm, I guess because $-1$ is an element of the quaternion group so any combination $a\mathbf{1}+(a-1)\mathbf{(-1)}$ is a solution as this is essentially 1. $\endgroup$ – zooby Jun 16 at 3:57
  • $\begingroup$ yes, I meant when projecting the group algebra onto the quaternions many combinations project onto 1. Because (-1) is an element of the group. $\endgroup$ – zooby Jun 16 at 4:00
  • $\begingroup$ No, that doesn't work. Try squaring it: it won't work for most values of $a$. $\endgroup$ – Eric Wofsey Jun 16 at 4:00

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