Groups and Subgroups

A subgroup of a group G is a subset of G that is closed under multiplication and the taking of inverses, and is non-empty. (Proving that this is equivalent to whatever definition you're working from is easy).

Now, there are some restrictions on what order a subgroup can have: specifically, a group can only have subgroups of orders dividing its own order (by Lagrange), so we're interested in subsets of order 1, 2, 3, 4, 6, 8, 12, and 24. Now, S4 is pretty unusual in that it has subgroups of every order on that list (most groups don't: in particular, A4 has no subgroup of order 6, though proving that in a non-tedious way is probably beyond the tools that you currently have available to you).

So, we'll start off with the easy ones: the only possible option for a subgroup of order 1 is the trivial subgroup, containing the identity and nothing else (this is the unique subgroup of order 1 in any group). Up at the other end, the only option for a subgroup of order 24 is the whole group (because we have only 24 elements to choose from). Again, any group is a subgroup of itself, trivially.
Now, onto the interesting ones: we'll start with the primes (2 and 3): since all groups of prime order are cyclic (if you haven't seen a proof of this already, it's a pretty easy exercise), we need to find elements of order 2 and 3: these are both pretty easy to do: swapping any pair of elements gives an element of order 2 (so in your notation, we'd take the {ABCD, ABDC}, for example, as our subgroup (obviously, you could pick any other pair of elements to swap and do the same, so we've actually got 6 subgroups of order 2). Similarly, we can get our subgroup of order 3 by just cycling three elements around (so we'd have {ABCD, ACDB, ADBC}, for example (and again, we can pick any other set of three, so there are 4 subgroups of order 3). We can also sneak in a subgroup of order 4 this way, by just cycling everything around (so we'd have

/r/askmath Thread