Let H be a subgroup of a group G. Prove that o(H) = o(x−1Hx), where x ∈G.

Answer 1

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.