In my undergraduate Group Theory class, while discussing the impossibility of equivalence relations on groups, my professor said that the set of all groups does not exist.

Are there any group subcategories for which the set of all such groups DO exist? As examples,

i) Does the set of all cyclic groups exist?

ii) Does the set of all finite groups exist?

iii) Does the set of all groups of order n, for some integer n, exist?

Thank you.

  • 2
    $\begingroup$ math.stackexchange.com/questions/226413/… $\endgroup$ – Asaf Karagila Nov 14 '12 at 19:00
  • $\begingroup$ More generally, a collection of groups being a set requires that there is a set $S$ such that the underlying set of every group in your collection is contained in this set. $\endgroup$ – Thomas Andrews Nov 14 '12 at 19:30
  • $\begingroup$ Tarksi-Grothendieck set theory is quite convenient because it allows you to work inside universes. The set of all groups in one universe is not an element of that universe, true - but, this set certainly exists, and its an element of the next universe. $\endgroup$ – goblin GONE May 31 '13 at 2:57

Yes and no. The underlying set of a trivial group can be any one-element set and the class of all one-element sets is a proper class (i.e. not a set). However, if we consider only groups up to isomorphism, then the set of cyclic groups exists (there is one for each natural numbre and there is $\mathbb Z$). In the same sense the set of groups of order $n$ (or more easily: The set of operations on a given set of $n$ elements such that this makes it a group) exists, and also the set of all finite groups.

  • 3
    $\begingroup$ To give an example, the group $S_\infty$ (the set of all permutations of $\Bbb N$ that move only finitely many elements) contains an isomorphic copy of every finite group, which is a corollary of Cayley's Theorem. To get a group that contains an isomorphic copy of every cyclic group, one can consider $S_\infty * \Bbb Z$. $\endgroup$ – user123641 May 23 '14 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.