Proving that H=S_4 for H subset of S_4

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Suppose that H is a subset of S_4, and that H contains (12) and (234). Prove that H=S_4.

The order of S_4 is 24.
The order of H is 2.

(12)(234) = (1342) shows that the finite subset of group H is closed under the group operation, so it is a subgroup of H.

Then, using Lagrange's Theorem, |S_4:H|=24/2 = 12. so the order of H is 12 by this.

And now I am lost on how I can prove that H = S_4.
 
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math_nerd said:
The order of H is 2.
why is it 2? how about looking at compositions of the multiplications and inverses that must be contained in H for it to be a group
 
Suppose that H is a subset of S_4, and that H contains (12) and (234). Prove that H=S_4.

The order of S_4 is 24.
(12)(234) = (1342) shows that the finite subset of group H is closed under the group operation, so it is a subgroup of H.
Then, using Lagrange's Theorem, |S_4:H|=24/2 = 12. so the order of H is 12 by this.

The previous post had a mistake. Sorry about that.

Now, to show that H is a subgroup of K: look at compositions of multiplications and inverses in H for it to be a group? How do I do this? I'm sorry this probably sounds really dumb, but all I know is that, aH=Ha iff H=aHa^-1 (property of cosets). I just don't see how this is done, and whatvthat will show.
 
does is say H is a subgroup?
 
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...

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