- 1,456
- 44
Let ##G## and ##H## be groups, and let ##\phi : G \to H## be an isomorphism. I want to show that ##\phi^{-1} : H \to G## is also an isomorphism. First, note that ##\phi^{-1}## is clearly a bijection as ##\phi## is its inverse. Second, let ##a,b \in H##. Since ##\phi## is surjective, there exist ##x,y \in G## s.t ##a = \phi(x)## and ##b = \phi(y)##. Then ##\phi^{-1}(ab) = \phi^{-1}(\phi(x) \phi(y)) = \phi^{-1}(\phi(xy)) = xy = \phi^{-1} (\phi (x)) \phi^{-1} (\phi(y)) = \phi^{-1}(a) \phi^{-1}(b)##.
Here is my question, why did I only have to use the fact that ##\phi## is surjective and a homomorphism in showing that ##\phi^{-1}## is a homomorphism? Why didn't I have to use that it is injective?
Here is my question, why did I only have to use the fact that ##\phi## is surjective and a homomorphism in showing that ##\phi^{-1}## is a homomorphism? Why didn't I have to use that it is injective?