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Question is in the title. For bonus points, construct the group $G$ such that it also has no infinite proper subgroups. (This second question relates to the Prüfer group, but that group is abelian, and clearly $G$ is nonabelian since it has nonabelian subgroups.)

Ignoring the second constraint for now, it is clear that the direct product of every finite group contains every finite group as a subgroup, but it is not a very "natural" group. Are there any examples of more common infinite groups that also happen to have every finite group as a subgroup?

Mario Carneiro
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1 Answers1

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Pick one representative for each isomorphism class of finite group —there are countably many of these. Now build the direct product of this countable family. You can also take only the symmetric groups, if you want, and there are many other variations.

The symmetric group on infinitely many letters is another example, or the restricted symmetric group on infinitely many letters (that is, the subgroup of the former of permutations which move finitely maany letters)

Mariano Suárez-Álvarez
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    The first one is my example, but the symmetric group example is very nice. But neither one satisfies the second criterion. Is there a reason to believe that it can't be satisfied? – Mario Carneiro Jan 07 '14 at 21:04
  • I only repeated your construction to note that one can actually do it with fewer groups: take only the symmetric groups, or the $GL$s or the altenrating groups of... – Mariano Suárez-Álvarez Jan 07 '14 at 21:06