Let \(\mathbb F\) be a field of characteristic \(p\), and \(G\) a group of order \(p^n\). Let \(R=\mathbb F[G]\) be the group ring (group algebra) of \(G\) over \(\mathbb F\), and let \(u:=\sum_{x\in G}x\) (so \(u\) is an element of \(R\)).
Prove that \(u\) lies in the center of \(R\).
Verify that \(Ru\) is a 2-sided ideal of \(R\).
Show there exists a positive integer \(k\) such that \(u^k=0\). Conclude that for such a \(k\), \((Ru)^k=0\).
Show that \(R\) is not a semi-simple ring.
Warning: Please use the definition of a semi-simple ring: do not use the result that a finite length ring fails to be semisimple if and only if it has a non-zero nilpotent ideal.
Solution
1. The element\[ u=\sum_{x\in G}x \]lies in the center of \(R=\mathbb F[G]\).
For any \(g\in G\), \[ gu=\sum_{x\in G}gx=\sum_{y\in G}y=u \] because left multiplication by \(g\) permutes the elements of \(G\). Similarly, \[ ug=\sum_{x\in G}xg=u. \] Thus \(gu=ug\) for every basis element \(g\in G\), and therefore \(u\) commutes with every element of \(R\).
2. One has\[ Ru=\mathbb F u, \]and \(Ru\) is a nonzero two-sided ideal of \(R\).
If \[ r=\sum_{g\in G}a_g g, \] then by 1, \[ ru=\sum_{g\in G}a_g(gu)=\left(\sum_{g\in G}a_g\right)u. \] Hence \(Ru\subseteq\mathbb F u\), while the reverse inclusion is immediate since \(\mathbb F\subseteq R\). Thus \(Ru=\mathbb F u\). Because the group elements form an \(\mathbb F\)-basis of \(R\), the element \(u\) is nonzero. Finally, centrality of \(u\) gives \[ (Ru)R=R(uR)=R(Ru)\subseteq Ru, \] so \(Ru\) is two-sided.
3. The element \(u\) is square-zero:\[ u^2=0. \]Consequently\[ (Ru)^2=0. \]
Using \(xu=u\) for every \(x\in G\), \[ u^2=\left(\sum_{x\in G}x\right)u =\sum_{x\in G}u =|G|u=p^n u=0 \] because \(\operatorname{char}\mathbb F=p\). Thus one may take \(k=2\). Since \(Ru=\mathbb F u\) by 2, every product of two elements of \(Ru\) is a scalar multiple of \(u^2\), hence is zero.
4. The ring \(R\) is not semisimple.
Assume for contradiction that \(R\) is semisimple as a left \(R\)-module. Then every submodule of the semisimple module \(R\) is a direct summand, so there is a left ideal \(J\) such that \[ R=Ru\oplus J. \] Let \[ \pi:R\longrightarrow Ru \] be the projection along \(J\). This map is left \(R\)-linear. Put \[ e=\pi(1)\in Ru. \] For every \(r\in R\), \[ \pi(r)=\pi(r\cdot1)=r\pi(1)=re. \] Since \(\pi\) restricts to the identity on \(Ru\), in particular \(\pi(e)=e\). Hence \[ e^2=\pi(e)=e. \] But \(e\in Ru\) and \((Ru)^2=0\) by 3, so \(e^2=0\). Thus \(e=0\), which would imply \[ \pi(r)=re=0 \] for every \(r\in R\), contradicting that \(\pi\) is the identity on the nonzero submodule \(Ru\). Therefore \(R\) is not semisimple.