# Permutation Group

1. Dec 4, 2009

### carmine

1. The problem statement, all variables and given/known data

Suppose G is a group generated
by the two permutations (1 2 3 4 5) and
(1 2)(3 4). Decide which group G is and

2. Relevant equations

3. The attempt at a solution
So i crunched this out and I found

identity
15->2 cycles
20->3 cycles
and 24->5 cycles

So i think this is the A5.

Is there an easier way to determine this, other than crunching out the numbers.

2. Dec 5, 2009

### lanedance

how many elements do you think are in your group?

3. Dec 5, 2009

### carmine

I get 60 elements....took a while, I'm just assuming there is an easier way..

4. Dec 5, 2009

### Dick

Yes, there's somewhat easier ways. Both your generators are even permutations, so you know G will be a subgroup of A5. Now use Lagrange's theorem. Your first one has order 5, so you know 5 divides the order of G. The second has order 2. Now can you find subgroups of order 3 and 4? That would do it, right?

Last edited: Dec 5, 2009
5. Dec 5, 2009

### latentcorpse

wow. i've forgotten lots of group theory since last year. why is (12345) even? isn't (12345)=(12)(23)(34)(45)(51) i.e. the product of an odd no. of transpositions and thus the whole permutation is odd?

even using lagrange's theorem,
so if 5 divides 60 then there's a subgroup of size 12 and if 2 divides 60 then there's a subgroup of size 30, no?

6. Dec 5, 2009

### Dick

You've forgotten how to multiply permutations, that's for sure. (12345)=(15)(14)(13)(12). That's an even number of transpositions. Your product comes out to be (2345). And not necessarily for the second part. The order of a subgroup divides the order of the group. That's all.

7. Dec 5, 2009

### latentcorpse

ok. checked my notes.
a k-cycle is even if k is odd as it can be written as a product of k-1 transpositions
(12345) is a 5-cycle and is therefore even as 5 is odd i.e. it can be written as (12)(23)(34)(45)
the general case given in my notes was (x1 x2 x3 ... xk)=(x1 x2)(x2 x3) ... (xk-1 xk) so i just fitted that to this case.

so the order of a subgroup divides the group. generators are always subgroups. so we know G is divisible by 5 and by 2. so #G is a multiple of 10. how can we deduce anything more?

8. Dec 5, 2009

### Dick

As I told carmine. Start taking products of the generators. See if you can find a subgroup of orders 3 and 4.

9. Dec 5, 2009

### latentcorpse

i get (12)(12345)=(1543) which is of order 4
and (12)(34)(12345)=(153) which is of order 3

so that tells us #G is divisible by 2,3,4,5

2x3x4x5=120
but 60 is also divisible by 2,3,4,5
nothing smaller than 60 is divisible by 2,3,4,5
so G=A5
yeah?

10. Dec 5, 2009

### Dick

The second one looks good. (1543) isn't in the group generated by the two given elements. It's an odd permutation. Notice, I didn't say 'find an element of order 4'. You aren't going to find one. I said 'find a subgroup of order 4'. It doesn't have to be cyclic.

11. Dec 5, 2009

### latentcorpse

subgroup of order 4 is $H=\{ (153),(145),(135),(235) \}$

so now we can use lagrange to say 4 divides the size of the group.

is the argument at the end of my last post ok for showing that G=A5?

12. Dec 5, 2009

### Dick

That's NOT a subgroup! It doesn't even have the identity in it. Think about it. A subgroup of order 4 is going to have to be generated by two elements of order 2. Right, the argument is fine. You just need to show the order of G is divisible by 4.

13. Dec 5, 2009

### latentcorpse

depressing to forget identity

so $H=\{ e , (12) , (34) , (12)(34) \}$ would do the trick.

14. Dec 5, 2009

### Dick

At least you've got a group there. But again (12) and (34) aren't in G. They are ODD. So it's not a SUBgroup.

15. Dec 5, 2009

### latentcorpse

ok (12345)(12)(34)=(135)
(34)(12345)(12)=(145)
(12)(12345)(34)=(235)

so H={(135),(145),(235)} looked ok until i tried combinations and found it wasn't closed under group action. any suggestions?

16. Dec 5, 2009

### Dick

Find the group generated by the two elements (12)(34) and (14)(32). What's it's order? You might ask how I knew (14)(32) was in G. Good question. I believe that carmine was correct in concluding the original two elements generate A5. So carmine knows how to check that (14)(32) is in G. I didn't bother to figure out how to generate it. Can you?

Last edited: Dec 5, 2009
17. Dec 5, 2009

### latentcorpse

ok well (12)(34)(14)(32)=(13)(24)

so H={e,(12)(34),(13)(24),(14)(23)} which is my subgroup of order 4 and then i can use my argument to deduce G=A5, yes?

so all that remains to be proven is that (14)(32) is in G

i assume i just have to play about with combinations of the cycles
(12)(34) and (12345) till it works?

also, does lagrange's theorem work on elements of size 3 as well as subgroups of order 3? i didn't find a subgroup of size 3, all i found was an element of order 3.

thanks for the revision tutorial in group theory. will be useful for my course in lie groups next semester hopefully!

Last edited: Dec 5, 2009
18. Dec 5, 2009

### Dick

Yes, you just have to play around with it until you find one. If you didn't find a subgroup of order 3 you should probably find one. That should be easy. An element of order 3 will generate it. Good luck with Lie groups. That's a whole nother ball of wax. But if you didn't understand finite groups, Lie groups would be hopeless.

19. Dec 6, 2009

### latentcorpse

rewinding about 7 months i was actually pretty competent at this stuff - clearly my christmas will need to involve some revision!

so the subgroup of order 3 will be {e,(153),(135)}

20. Dec 6, 2009

### Dick

Yes, that's one of them.