Can anybody help me in finding the all possible subgroups of the group

<x,y:x²=y³=(xy)¹²=1>. Actually i want to find the the number of all subgroups of this group.

Results 1 to 4 of 4

- December 13th 2010, 05:36 AM #1

- Joined
- Dec 2008
- Posts
- 12

- December 13th 2010, 10:38 AM #2

- Joined
- Oct 2009
- Posts
- 4,261
- Thanks
- 2

Are you sure you meant the above or perhaps you meant to give a presentation of a group by generators and relators, in the

form of ?

Because if you meant what you wrote and not the above then we get:

, from where we get

that your group can be presented as ... Check this.

Tonio

- December 14th 2010, 07:59 PM #3

- Joined
- Nov 2010
- Posts
- 193

- December 15th 2010, 01:14 AM #4
I don't know what he is doing either - certainly he has done something wrong in his calculations (his first group abelinises to while the second abelianises to , by Tietze transformations...)

I suspect he didn't see the `=1' in your presentation, but even then your group abelianises to . So...yeah, he's done something wrong...I have not idea where he gets his final presentation from.

Anyway, to address your question, there isn't really a `neat' way of finding all the subgroups of a group that I know of. You just need to do it by hand. So, what is the order of your group? Which groups have order dividing your groups order?

Next, take elements from your group and see what (sub)groups they generate. Then you want to write the subgroups out in a lattice, much like you can find here.