Proving Conjugacy of Subgroups in a G-set X

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In summary, the conversation discusses the concepts of G-sets and conjugate subgroups. It is shown that if x is an element of a G-set X and b is an element of the group G, then S(bx) is equivalent to bS(x)b^-1. It is also shown that if S(x) and S(y) are conjugate subgroups, then |Gx| = |Gy|. However, the attempt at a solution is incorrect as it assumes that abx=x implies bx=x, which is not necessarily true. The correct approach would be to start with b^-1abx = x and continue from there.
  • #1
hsong9
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Homework Statement


Let X be a G-set and let x and y denoted elements of X.
a) If x in X and b in G, show that S(bx) = bS(x)b-1
b) If S(x) and S(y) are conjugate subgroups, show that |Gx| = |Gy|


The Attempt at a Solution


Let S(x) = {a in G | ax=x}
Let S(bx) = {a in G | abx=x} => abx = x => bx = x => bxb-1 = xb-1

bS(x)b-1 = baxb-1 = bxb-1 = xb-1

Thus S(bx) = bS(x)b-1

b) Since S(x) and S(y) are conjugate subgroups, |S(x)| = |S(y)|
so by |Gx|=|G:S(x)|, |Gx|=|Gy|
correct?
 
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  • #2
hsong9 said:
Let S(x) = {a in G | ax=x}
Let S(bx) = {a in G | abx=x}
=> abx = x
=> bx = x
=> bxb-1 = xb-1

No, that does not hold. That abx=x does not imply that bx=x, as that would mean that b stabilizes x which is certainly not (necessarily) true.



bS(x)b-1 = baxb-1 = bxb-1 = xb-1

Whoa, there. You've now equated a group with an element in a group.

Start again with b^-1abx = x and see what you can say.
 

1. What is the definition of a conjugate subgroup in a G-set X?

A conjugate subgroup in a G-set X is a subgroup that is obtained by applying a group element from the larger group G to the subgroup in question. This results in a new subgroup that has the same structure as the original subgroup, but with a different set of elements.

2. How do you prove that two subgroups in a G-set X are conjugate?

To prove that two subgroups in a G-set X are conjugate, you must show that there exists a group element from G that can map one subgroup to the other. This can be done by explicitly finding the group element or by showing that the two subgroups have the same structure and therefore must be conjugate.

3. What is the significance of proving conjugacy of subgroups in a G-set X?

Proving conjugacy of subgroups in a G-set X is important because it allows us to establish a relationship between two subgroups and understand how they are related to each other within the larger group G. This can provide insights into the structure of the group and help solve problems in group theory.

4. Can two subgroups in a G-set X be conjugate if they are not isomorphic?

Yes, two subgroups can be conjugate in a G-set X even if they are not isomorphic. This is because conjugacy is a relationship between subgroups that is based on the structure and elements within the subgroups, not on their isomorphism.

5. How can the concept of conjugacy be applied in real-world situations?

The concept of conjugacy in group theory has many applications in fields such as chemistry, physics, and computer science. For example, in chemistry, conjugate subgroups can help determine the symmetry of molecules and predict their properties. In physics, conjugacy can be used to analyze the symmetries of physical systems. In computer science, conjugate subgroups can be applied in cryptography to generate secret keys for secure communication.

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