Let φ: G → G be an isomorphism from the group G to the group G. Since φ is bijective, its inverse ψ: G → G exists according to Theorem 1.9.1, which states that a function f: X → Y has an inverse if and only if it is bijective. To prove that ψ is an isomorphism, it suffices to show that ψ is bijective.
Let - G G be an isomorphism from the group G to the group G- In part.docx
1. Let : G G be an isomorphism from the group G to the group G. In particular, is bijective so
that its inverse 1 : G G exists (see 1.9.1). Prove that 1 is an isomorphism.
1.9.1 Theorem. A function f : X Y has an inverse if and only if it is bijective.
Proof. Letf:XY beafunction.
() Assume that f has an inverse f 1
. (f injective?) Let x,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 and surjective, it is
bijective.
() Assume that f is bijective. Define f 1
: Y X by letting 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 is injective, there is at most one
such x.) For x X, we have f 1
(f(x)) = x (since f 1
(f(x)) is defined to be 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 element that f sends to y).
Therefore, f 1
is an inverse of f.
Solution
You mulitply and divide and your answer shall me y = 384.4
