Let H be a subgroup of a group G. Prove that o(H) = o(x−1Hx), where x ∈G.
Let o(H) = n. Then H has n elements.
Let y_1, y_2, ..., y_n be the elements of H.
Consider the set { x^(-1) y_1 x, x^(-1) y_2 x, ..., x^(-1) y_n x }.
We claim that this set is equal to x^(-1) H x.
Indeed, if x^(-1) h x is in x^(-1) H x, then h is in H by definition of the subgroup. Therefore, there exists a y_i in H such that h = y_i. Thus x^(-1) h x = x^(-1) y_i x is in our set.
Conversely, if x^(-1) y x is in our set, then there exists an element h in H such that x^(-1) y x = x^(-1) h x. Multiplying by x on both sides, we get y = h, which is in H. Therefore, x^(-1) y x is in x^(-1) H x.
Thus, we've shown that { x^(-1) y_1 x, x^(-1) y_2 x, ..., x^(-1) y_n x } = x^(-1) H x.
Now, we claim that the elements in this set are distinct.
Suppose x^(-1) y_i x = x^(-1) y_j x for some i and j. Then y_i = x x^(-1) y_i x x^(-1) = x x^(-1) y_j x x^(-1) = y_j. This contradicts the fact that the y_i's are distinct elements of H. Therefore, the elements in the set { x^(-1) y_1 x, x^(-1) y_2 x, ..., x^(-1) y_n x } are distinct.
So the set x^(-1) H x has n elements.
Therefore, o(x^(-1) H x) = n = o(H), as desired.
To prove that the order of the subgroup H is equal to the order of the conjugate subgroup x^(-1)Hx, where x is an element of the group G, we need to show that every element in H is mapped to a distinct element in the conjugate subgroup.
Let's start by assuming h1 and h2 are two distinct elements in H. We want to show that their conjugates, x^(-1)h1x and x^(-1)h2x, are also distinct.
Suppose, for the sake of contradiction, that x^(-1)h1x = x^(-1)h2x. Then, multiplying both sides of the equation by x from the left gives us:
x(x^(-1)h1x) = x(x^(-1)h2x)
This simplifies to:
(h1)(x(x^(-1))h2x) = (h1)(h2)
Since the group operation is associative, we can rearrange the expression as:
h1[(x(x^(-1)))h2]x = h1(h2)
Using the fact that x(x^(-1)) is the identity element e, we have:
h1(ex) = h1(h2)
This further simplifies to:
h1x = h2
Since h1 and h2 are distinct elements in H, we have arrived at a contradiction. Therefore, if h1 and h2 are distinct elements in H, their conjugates x^(-1)h1x and x^(-1)h2x must also be distinct in x^(-1)Hx.
As a result, every element in the subgroup H is uniquely mapped to an element in the conjugate subgroup x^(-1)Hx. Therefore, the number of elements in H is equal to the number of elements in x^(-1)Hx, leading to o(H) = o(x^(-1)Hx).