Thread: Symmetric Group Subgroups

and how did you arrive at that conclusion? it's not that i don't believe you, but rather, in mathematics, the pudding is in the proof.

I tested every element and saw gxg^-1 was always in <x>. Then I stopped because you said there was only 1! But I see that I could find counterexamples for the other subgroups to show they are not.

What I still don't really understand is how to find the subgroups in the first place.

this is where lagrange's theorem comes in. besides S3 and 1, you can only have subgroups of order 2 and 3.

every element of order 2 generates a subgroup of order 2. S3 has 3 elements of order 2, hence 3 subgroups of order 2. see?

every element of order 3 generates a subgroup of order 3. S3 has two elements of order 3, both of which generate the same subgroup.

now, 2 and 3 are prime, so subgroups of prime order can only be made from a generator. we've found all the generators, so we've found all the subgroups

(this is no longer true in larger groups, be thankful S3 is so nice).

Looks like I need to go study up on that theorem. Thanks a lot for all your help! You cleared up a lot for me.