Proof that S is a generating set.

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Homework Statement


I have recently started a new course in Algebra. I have to proof that if S is a subset of a finite group G, with an order greater than half the order of G, S is a generating subset for G.

Homework Equations


I haven't had cosets nor Lagrange theorem so I suppose I should try to prove it from scratch.

The Attempt at a Solution


I have put my mind to the problem for a while now but didnt come up with any meaningful clues. I would be very thankful for any tips.
 
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kasperrepsak said:

Homework Statement


I have recently started a new course in Algebra. I have to proof that if S is a subset of a finite group G, with an order greater than half the order of G, S is a generating subset for G.

Homework Equations


I haven't had cosets nor Lagrange theorem so I suppose I should try to prove it from scratch.

The Attempt at a Solution


I have put my mind to the problem for a while now but didnt come up with any meaningful clues. I would be very thankful for any tips.

Let H be the group generated by S. That's a subgroup of G and it has at least as many elements as S. Suppose H isn't the whole group G. Then there is an element x of G that's not in H. Can you show H and xH are disjoint sets and they have the same number of elements? Sure, xH is a coset. But you don't have to know that to do the proof.
 
Last edited:
Dick said:
Let H be the group generated by S. That's a subgroup of G and it has at least as many elements as S. Suppose H isn't the whole group G. Then there is an element x of that's not in H. Can you show H and xH are disjoint sets? Sure, xH is a coset. But you don't have to know that to do the proof.

Thanks for the fast reply ! : ) Thank you I will work with that.
 
Ok so I know that H and xH must be disjoined sets (easy to proof), they have the same number of elements and both must be in G. But since the order of H is greater than half the order of G, H and xH unified would have more elements than G which is a contradiction. Therefore H must be the whole group G. Is this right?
 
kasperrepsak said:
Ok so I know that H and xH must be disjoined sets (easy to proof). H and xH have the same number of elements. They both must be in G. But since the order of H is greater than half the order of G, H and xH combined would have more elements than G which isn't possible. Therefore H must be the whole group G. Is this right?

Absolutely correct.
 
Ok thanks again for your help : ).
 
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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