Let be any nonempty set and let be a group for each . The restricted direct product or direct sum of the groups is the subset of elements such that for all but a finite subset .
- Prove that is a subgroup of .
- Prove that is normal in .
- Let and let be the th integer prime. Show that if for all , then every element of the restricted direct product of the has finite order but that has elements of infinite order. Show that in this example the restricted direct product is the torsion subgroup of the direct product. (See this previous exercise.)
- Note that , where we may take . Thus is nonempty. Now suppose and are in via the finite subsets . Note that is finite, and that if , then . Thus via . By the Subgroup Criterion, .
- Let via the finite set and let . Now . Note that for , . Thus via ; hence is normal in .
- Let be in the restricted direct product via a finite set . Define , where this product is simply multiplication in the integers. (Since is finite, this product exists.) Note that if since divides , and if since . Thus , and has finite order.
On the other hand, consider in the direct product defined by . Now for all odd primes , generates , so that has order . Then cannot have finite order, since for all integers , some component of is not 0; for instance, take any such that .
Now the first part of this problem showed that the restricted direct product is contained in the torsion subgroup. Suppose now that is in the torsion subgroup of . Then for some integer , we have ; hence for all . Now if and , then since and has order . Thus for all , we have . In particular, for all but finitely many , namely the set . Thus is in the restricted direct product, and in this case the restricted direct product and the torsion subgroup coincide.