✔ 最佳答案
Let set S = {x ∈ G | x^2 = e}
1 e * e = e => e ∈ S
2 If Let x,y ∈ S, then x^2 = y^2 = e
(xy)^2 = xyxy = xxyy = e * e = e
xy ∈ S
3 x^2 = e => x = x^(-1)
So, [x^(-1)]^2 = e => x^(-1) ∈ S
From (1) - (3), we conclude that S is nonempty and closed under products and inverses. So, S is a subgroup of G.