Let $R$ be a ring and $G$ an infinite group. Prove that $R(G)$ (group ring) is not semisimple.

My idea was to suppose it is semisimple, then $R(G)$ is left artinian and $J(R(G))=0$. I was trying to make a ascending chain of ideals that won't stop, then it is not left noetherian, by Hopkins theorem it is not left artinian, a contradiction. I also tried to make a descending chain that won't stop, so it is not left artinian, but I wasn't successful. So please help me.

  • $\begingroup$ Your trying it seems to use only the finiteness (artinian or noetherian) properties of $R(G)$. This makes me think that you are not on the right track. $\endgroup$
    – user89712
    Dec 25, 2013 at 13:30
  • $\begingroup$ actually I don't know what should I do,what is your Idea? $\endgroup$
    – kpax
    Dec 25, 2013 at 13:50
  • $\begingroup$ Yeah, I think user's feeling is correct. "R[G] artinian implies G finite" is a nontrivial proof appearing in Lambek's algebra book and Connell's thesis. $\endgroup$
    – rschwieb
    Dec 25, 2013 at 14:59
  • $\begingroup$ Are you trying to show "semisimple" in the sense that the Jacobson radical is zero, or semisimple in the sense that it is a direct sum of simple rings? $\endgroup$
    – user2055
    Dec 25, 2013 at 15:15
  • $\begingroup$ @ Jason Polak ,I must show that it is not semisimple,so I suppose it is,and I wanted to show it comes to a contradiction. $\endgroup$
    – kpax
    Dec 25, 2013 at 15:26

1 Answer 1


Comments on the partial progress.

Unfortunately, this is a case of something that sometimes happens with students: a strong condition was given (semisimple Artinian) and then you decided to quickly drop down to something weaker (Artinian/Noetherian) possibly because you feel more comfortable with that condition.

It is possible to prove that if $G$ is infinite then $R[G]$ can't be Artinian, but as I alluded to in the comments it's not easy enough to be a homework problem.

If you start with assuming $G$ is infinite and then try to show that $R[G]$ isn't Noetherian, you are unfortunately doomed to failure, because (I think I am recalling correctly) there are examples of infinite groups $G$ such that $R[G]$ is Noetherian.

This highlights some of the dangers of "simplifying" givens too quickly. We can prove it in the following way taking full advantage of supposed semisimplicity of $R[G]$. Specifically, we'll use the fact that ideals are summands.


  1. Let $A$ be the augmentation ideal of $R[G]$. That is, it is the kernel of the projection $R[G]\to R$ which sends $g\mapsto 1$ for all $g\in G$. Show that $1-g\in A$ for all $g\in G$.
  2. Suppose $R[G]$ is semisimple. Then $A$ is a summand of $R[G]$, and in particular we can find an idempotent $e\in R$ such that $R[G]e=A$ and $R[G]e\oplus R[G]f=R[G]$ where $f=1-e$ is another idempotent. Prove that $(1-g)f=0$ for every $g\in G$ using the previous point. Consequently, $f=gf$ for all $g\in G$.
  3. $f$ has to be nonzero, since the augmentation ideal is proper. Let $h\in G$ be a term of $f$ that has a nonzero coefficient. Show using the previous point that $gh$ appears with a nonzero coefficient in the expression of $f$.
  4. Since every element of $G$ can be expressed as $gh$ for some $g$, this means that all elements of $G$ have nonzero coefficients in the expression of $f$... but it isn't possible for $f$ to have infinitely many nonzero terms. Thus $G$ isn't infinite.
  • $\begingroup$ Dear @user :Thanks, that was good inspiration for some additional comments about the partial progress. While it's possible to pursue the "suppose $G$ is infinite, now I'll show $R[G]$ isn't Artinian" idea, it would be excessively long for a solution I think. The "suppose $G$ is infinite, now I'll show $R[G]$ isn't Noetherian" idea just doesn't work (not without backtracking and picking up more conditions from semisimplicity.) $\endgroup$
    – rschwieb
    Dec 25, 2013 at 19:33
  • $\begingroup$ thank you for your complete explanation,and also happy new year,I wish the best for all. $\endgroup$
    – kpax
    Dec 25, 2013 at 20:41
  • $\begingroup$ @kpax Yeah, Happy Holidays to you too! $\endgroup$
    – rschwieb
    Dec 25, 2013 at 22:47
  • $\begingroup$ Dear @rschwieb : In working through your suggested approach, I became confused about the existence of the idempotent $e \in R$ as opposed to $e \in R[G]$. My understanding is that semisimplicty only guarantees the latter in this case? Everything else seems to work the same regardless of which is the case, so I am wondering if that is a typo, or there is something I am missing. Thanks $\endgroup$
    – user1348
    Sep 8, 2016 at 18:33
  • $\begingroup$ @user1348 Hi, yes, I think you're right that I left out the $[G]$ by accident. Thanks for letting me know! $\endgroup$
    – rschwieb
    Sep 8, 2016 at 18:53

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