All Subgroups of S3: Lagrange's Theorem Explained

  • Thread starter Thread starter e179285
  • Start date Start date
  • Tags Tags
    Subgroup
e179285
Messages
24
Reaction score
0
The question wants all subgroups of S3 . If H≤S 3 , then ; IHI=1,2,3,6 by Lagrance's Theorem.

In other words, order of H can be 1,2,3 and 6.

What ı want to ask is how to write subgroup of S3. For example,is H 1 (1) ?
 
Physics news on Phys.org
For "H_1", yes, any subgroup must contain the identity so if H contains only one member, it must be just the identity permutation, (1).

There are, in fact, 3 different subgroups of order 2: {e, (12)}, {e, (13)}, and {e, (23)}. There is the single subgroup of order 3: {e, (123), (132)}. Of course, the subgroup of order 6 is the entire group.
 
HallsofIvy said:
For "H_1", yes, any subgroup must contain the identity so if H contains only one member, it must be just the identity permutation, (1).

There are, in fact, 3 different subgroups of order 2: {e, (12)}, {e, (13)}, and {e, (23)}. There is the single subgroup of order 3: {e, (123), (132)}. Of course, the subgroup of order 6 is the entire group.

What can H5 be?
 
? You just said that every subgroup of S3 (every subgroup of any group of order 6) must have order 1, 2, 3, or 6 (a divisor of 6). What do you mean by "H5"?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top