Abstract Algebra/Group Theory/Homomorphism/Homomorphism Maps Inverse to Inverse

Theorem

Let f be a homomorphism from group G to Group K.

Let g be any element of G.

f(g^-1) = [f(g)]^-1

Proof

0.   $f(g)⊛f(g^{-1})=f(g\ast g^{-1})$	f is a [[../Definition of Homomorphism, Kernel, and Image|homomorphism]]
1.   $f(g)⊛f(g^{-1})=f(e_G)$	definition of [[../../Group/Definition of a Group/Definition of Inverse|inverse]] in G
.
2.   $f(g)⊛f(g^{-1})=e_K$	homomorphism f [[../Homomorphism Maps Identity to Identity|maps identity to identity]]
3.   $[f(g){]}^{-1}⊛f(g)⊛f(g^{-1})=[f(g){]}^{-1}⊛e_K$	as f(g) is in K, so is its [[../../Group/Definition of a Group/Definition of Inverse|inverse]] [f(g)]−1
.
4.   $e_K⊛f(g^{-1})=[f(g){]}^{-1}$	[[../../Group/Definition of a Group/Definition of Inverse|inverse]] on K, eK is [[../../Group/Definition of a Group/Definition of Identity|identity]] of K
5.   $f(g^{-1})=[f(g){]}^{-1}$	eK is [[../../Group/Definition of a Group/Definition of Identity|identity]] of K