Group G is A5: How to Determine the Group with Permutations

[carmine]

Homework Statement

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
prove your answer.

Homework Equations

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.

 

[lanedance]

how many elements do you think are in your group?

 

[carmine]

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

 

[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?

 

[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?

 

[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.

 

[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?

 

[Dick]

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

 

[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?

 

[Dick]

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?

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.

 

[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?

 

[Dick]

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?

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.

 

[latentcorpse]

depressing to forget identity

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

 

[Dick]

depressing to forget identity

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

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.

 

[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?

 

[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?

 

[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!

 

[Dick]

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!

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.

 

[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)}

 

[Dick]

Yes, that's one of them.

 

[latentcorpse]

how would one go about finding the number of subgroups of order 3?

 

[Dick]

how would one go about finding the number of subgroups of order 3?

You tell me this time. Give it some thought.

 

[latentcorpse]

there are 5 choose 3 = 20 different 3-cycles

but each subgroup of order 3 contains the identity and 2 of these 3-cycles

20/2=10 such subgroups?

 

[Dick]

You lucked out. 5 choose 3 is actually 10. Why is it still correct that there are 20 3-cycles?

 

[latentcorpse]

it's usually to do with some double counting trick. is it because 5 choose 3 gives us the number of ways of picking 3 things from 5 things but doesn't care about their order. i.e. it vies (135) and (153) as the same thing when in fact they are different?

 

[Dick]

it's usually to do with some double counting trick. is it because 5 choose 3 gives us the number of ways of picking 3 things from 5 things but doesn't care about their order. i.e. it vies (135) and (153) as the same thing when in fact they are different?

Exactly.