Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As you may know, in a finite direct product $\prod_{i\in I}S_i$ of finite nonabelian simple groups $S_i$, every normal subgroup is of the form $\prod_{i\in J,J\subseteq I}S_i$.

Normal subgroups of finite index in infinite product $\prod_{i\in I}S_i$ (with $I$ infinite) are still of the same form?

Now we consider the direct sum $\oplus_{i\in I}S_i$ with $I$ infinite. The question is the same.

What are normal subgroups of finite index in the infinite sum $\oplus_{i\in I}S_i$?

Thanks in advance.

share|cite|improve this question
Direct sum for groups? Do you want to mean free product instead? Or Abelian groups? – Berci Feb 27 '13 at 12:26
@Berci, it is still a common name for the coproduct. – Andreas Caranti Feb 27 '13 at 12:41
Ah, ok. So, it is free product. – Berci Feb 27 '13 at 12:54
Calling $\mathop{Supp} x = \{i\in I: x_i \ne 1\}$ the support of $x \in$ your infinite product/sum, take a look at the set $N_U$of all $x$ whose support is not an element of some fixed non-principal ultrafilter U on I. It's not too hard to show that $N_U$ is a normal subgroup. If all $S_i$ are isomorphic to each other, I think its index should be finite. – j.p. Feb 27 '13 at 13:31
I don't think he means the free product. He probably means the version of direct product in which only finitely many components of a group element are allowed to be nontrivial. – Derek Holt Feb 27 '13 at 14:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.