Question

Let φ : G → G′ be an isomorphism from the group G to the groupG′. In particular, φ is bijective so that its inverse φ−1 : G′ → Gexists (see 1.9.1). Prove that φ−1 is anisomorphism.

1.9.1 Theorem. A function f : X → Y has an inverse if and only ifit is bijective.
Proof. Letf:X→Y beafunction.
(⇒) Assume that f has an inverse f^−1. (f injective?) Letx,x′ ∈ X and assume that f(x) = f(x′). Then x =f^−1(f(x)) = f^−1(f(x′)) = x′.

Therefore, f is injective. (f surjective?) Let y ∈ Y . Put x =f^−1(y). Then x ∈ X and
f(x) = f(f^−1(y)) = y.

Therefore, f is surjective. Since f is both injective andsurjective, it is
bijective.
(⇐) Assume that f is bijective. Define f^−1 : Y → X byletting f^−1(y) be the unique x in X for which f(x) = y.(Since f is surjective, there is at least one such x and since f isinjective, there is at most one such x.) For x ∈ X, we havef^−1(f(x)) = x (since f^−1(f(x)) is defined tobe the element that f sends to f(x)). Similarly, for y ∈ Y ,f(f^−1(y)) = y (since f−1(y) is defined to be the elementthat f sends to y). Therefore, f^−1 is an inverse off.