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Subgroups of Alternating Group

by Kalinka35
Tags: group theory, permutation, subgroups
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Kalinka35
#1
Mar17-09, 08:22 PM
P: 50
Does A5 (the alternating group of degree 5) contain a subgroup of order m for each factor m of 60?


My intuition says yes, but I can't seem to find a way to prove this, short of writing out example subgroups for each factor, which is really tedious. Although if I have subgroups of orders 2, 3, 4, and 5 would it be enough to look at products of these?
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Daettil
#2
Mar17-09, 11:03 PM
P: 6
Perhaps you should try to construct a subgroup of order 30. Such a subgroup is normal so you can constuct one by taking the union of conjugacy classes...
Kalinka35
#3
Mar17-09, 11:40 PM
P: 50
How do you know that a subgroup of order 30 must be normal?
I know that A5 is simple so cannot contain a normal subgroup except for {e} and {A5}.

sutupidmath
#4
Mar18-09, 02:36 AM
P: 1,633
Subgroups of Alternating Group

There is a theorem that says: Let H be a subgroup of order t in a group G of order 2t. Then: H is normal in G, and moreover G/H={H,K}, where K consists of the t elements of G not in H.

Proof: Let g be an element of G such that g is in H. Then gH (the left coset) consists of exactly t elements as well. Moreover, since H is a subgroup=>gH=H=Hg.

Now, let r be any other element such that r is not in H. Then rH and H are going to be disjoint cosets(by another proposition: Two cosets are either identical or disjoint). But, again, rH must have exactly t elements, but in this case these t elements are the ones not contained in H, but rather in K. Now, the union of such cosets should give us the group G itself.

G={gH,rH}={Hg,Hr}=>gH=Hg => H<G (H is normal in G)

Edit: Another part of this theorem which probbably would help you prove what you want to prove is that: for every element g of G => g^2 is in H.

What i would probbably try to do is first determine how many three cycle permutations i.e. (abc) we have in A_5, and then how many 5-cycle, how many of the form (ab)(cd) etc. Then, this would give you an idea, say if there were a subgroup H of order 6, then all 3-cycles should be in H. Now, if the number of 3-cycles is greater than 6, then this would tell you that there is no such group of ord 6. So, try to work sth along these lines.

Edit2: As a matter of fact, from the top of my head i know that A5 does not have any normal subgroups of index 2,3 or 5. This is because,( i don't know whether you have been introduced yet), the group A_5 is not solvable.

In this case, since if a subgroup of ord 30 exists, then it must be normal, then if it is normal its index in A5 is 2, but since there is no normal subgroup of index 2 in A5, we conclude that there is no subgroup of ord 30 in A5.

Do you follow? (You still need fo fill in the 'why's?" though...
matt grime
#5
Mar18-09, 02:43 PM
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Quote Quote by sutupidmath View Post
Edit2: As a matter of fact, from the top of my head i know that A5 does not have any normal subgroups of index 2,3 or 5. This is because,( i don't know whether you have been introduced yet), the group A_5 is not solvable.
This is an odd statement. A_5 does not have a non-trivial normal subgroup of *any* index, since it is simple (not just not solvable - A_5 x C_2 is not solvable but has 2 normal subgroups, one of which has index 2). Why single out 2,3, and 5?

In this case, since if a subgroup of ord 30 exists, then it must be normal, then if it is normal its index in A5 is 2, but since there is no normal subgroup of index 2 in A5, we conclude that there is no subgroup of ord 30 in A5.
Again, this is odd. There are no normal subgroups - the central two steps of logic in this deduction are redundant, i.e. you can stop at


"In this case, since if a subgroup of ord 30 exists, then it must be normal <snip>. We conclude that there is no subgroup of ord 30 in A5"
sutupidmath
#6
Mar18-09, 11:03 PM
P: 1,633
Quote Quote by matt grime View Post
This is an odd statement. A_5 does not have a non-trivial normal subgroup of *any* index, since it is simple.Why single out 2,3, and 5?


Yes, you are right, of course!

When i pointed out normal subgroups of index 2,3,5, what i really had in mind was the solvability issue of A5(which by the way is not that relevant here). In other words, the fact that if A_5 would be solvable, then the only normal subgroups with index prime of A5, would respectively have indexes 2,3 or 5.

And, yes, my reasoning should have ended earlier, as you pointed out.
Kalinka35
#7
Mar18-09, 11:19 PM
P: 50
Taking a step back, how do I prove that A5 is simple? My approach would be to determine the conjugacy groups of A5 and their orders and use to these to show that there can be no other subgroups besides {e} and A5 itself. But I am not sure how to actually find the conjugacy groups. Or do you know of any other proof methods?
matt grime
#8
Mar19-09, 02:06 AM
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You omitted the word normal in your second sentence - you want to show that it has no subgroup of order 30, index 2, which is necessarily normal.

A_5 is in S_5. You know the conjugacy classes of elements in S_5 explicitly. Start from there: if two elements in A_5 are conjugate in S_5 when are they also conjugate in A_5? It's just a matter of calculations here.


So, do you know the conjugacy classes of elements in S_5?
Kalinka35
#9
Mar19-09, 11:55 AM
P: 50
Yes, yes, I did forget the word "normal."

So yes, I do have the conjugacy classes of S5 explicitly. I guess the key I'm missing is when two elements conjugate in S5 are also conjugate in A5. Is there a theorem about this?
matt grime
#10
Mar19-09, 01:07 PM
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Yes. But you can just work it out by doing it. Try S_3 and S_4 first. Or you can just look it up, of course. I can't decide if I condone that action in this case.
Kalinka35
#11
Mar19-09, 01:46 PM
P: 50
It's okay. I worked them out myself, but it just took a while so I was simply curious as to whether there was a simpler way.

Thanks.


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