Let and be groups and let be a group homomorphism. Prove that the image of is a subgroup of . Prove that if is injective then .
By a previous exercise, it suffices to show that is closed under multiplication and inversion. To that end, let . Then there exist such that and . Since , is closed under multiplication. Now let ; then there exists such that . Now , so that is closed under inversion. Thus is a subgroup of .
Suppose now that is injective. We can then defined a mapping by ; this is clearly a homomorphism and is bijective, hence an isomorphism. Thus .