Group Theory: Proof of the Formula for the Inverse of a Product

Group Theory: Proof of the Formula for the Inverse of a Product.
Given x, y in G, we prove carefully that (xy)^(-1) = y^(-1)x^(-1).

[Maurus]

I proved it in a different way:
First, xy is an element of G because G is closed under the group operation. Next, (xy)^-1 exists because G is a group and xy is in G.
Therefore, (xy)^-1 (xy) = e.
Multiply y^-1 on the right of both sides:
(xy)^-1 (xy) y^-1 = ey^-1
Apply associativity:
(xy)^-1 (x)yy^-1 = ey^-1
Y times it’s inverse is the identity. And any element times the identity is the element itself:
(xy)^-1 (x) = y^-1
Multiply x^-1 on the right of both sides:
(xy)^-1 (x)x^-1 = y^-1 x^-1
Apply the same reasoning as before, and:
(xy)^-1 = y^-1 x^-1

[divyadulmini]

l learned very much.Thank you very much sir ❤️❤️

[iHooKiDesign]

Why do you say (xy)^-1 = x^-1 y^-1 in the beginning but when you write it down saying you multiply the right part with the left one you suddenly only write (xy)(x^-1 y^-1). What happened with the -1 to the (xy), why is it not there anymore? Pls help

[NavidMoghaddas]

Could you prove the general case? a1, a2, ..., at ∈ G then (a1 ·a2···at)−1 = at^−1···a2^−1 ·a1^−1?

[ptyamin]

So this is the inverse of a product using *any* binary operation?

[newbek]

Question let (G, *) be a group prove that (x*y)^-1=x^-1*y^-1, for x, y ∈ G . is Solution