aguacate wrote:btilly - nice. I guess I don't see how (3456) generates the other cycles though. I haven't slept in a while.

Guess: is it because of the inverses? I don't have a solid understanding of what inverses look like yet.

Let me be painfully explicit.

(3456) does what? Maps 3 to 4, 4 to 5, 5 to 6 and 6 to 3.

(3456)(3456) does what? It maps 3 to 4 then 5. And 5 to 6 then 3. It also maps 4 to 5 then 6, and 6 to 3 then 4. So (3456)(3456) = (35)(46).

Multiply (35)(46)(3456) again and you'll find that 3 goes to 4 then 6, 6 goes to 3 then 5, 5 goes to 6 then 4 and 4 goes to 5 then 3. (I'm obviously putting the permutation to the left of the set being permuted, and applying the permutations right to left. Some people do it the other way.) So (35)(46)(3456) = (3654).

And (3654)(3456) gives you the identity permutation. As you'd hope because a 4 cycle, rotated 4 times should be the identity.

So that's how (3456) gave rise to those 4 permutations.

Similar calculations show you that (35)(3456) = (34)(56), (35)(35)(46) = (46), (35)(3654) = (36)(45), (35)I = (35). And, of course, applying (35) to those takes you back to the original.

Now we've showed that (35) applied to all 8 elements of our potential subgroup keep you in the subgroup. If we show that (3456) does as well then our subgroup is closed. (Because any member of the set is a product of things that take our subgroup back into our subgroup.) This calculation is likewise straightforward.

Now we add in (12). Since it commutes with all of the permutations that we've defined, it is easy to show that (3456), (35), and (12) form a subgroup of size 16. Which is the right size to be a Sylow-2 subgroup, and therefore is one.