# I Showing that inverse of an isomorphism is an isomorphism

#### Mr Davis 97

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?

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#### fresh_42

Mentor
2018 Award
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 $\phi^{-1}(ab) = \phi^{-1}(\phi(x) \phi(y)) = \phi^{-1}(\phi(xy)) \stackrel{(*)}{=} xy \stackrel{(*)}{=} \phi^{-1} (\phi (x)) \phi^{-1} (\phi(y)) = \phi^{-1}(a) \phi^{-1}(b)$.

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?
You did use it. Let's assume $\phi$ is only surjective and we have $\phi(u)=\phi(v)$. Then how could we justify $u=\phi^{-1}(\phi(u))=\phi^{-1}(\phi(v))=v$ which you used at the equations I marked with $(*)\,?$ So we actually used $\phi(u)=\phi(v) \Longrightarrow u=v$ which is precisely injectivity.

#### fresh_42

Mentor
2018 Award
One can define injectivity and surjectivity only by means of morphisms. Let's take the example above: $\phi\, : \,G \longrightarrow H$.

Then $\phi$ is injective, if and only if for any functions $\varphi , \psi \, : \, K \longrightarrow G$ from a set $K$ with $\phi \varphi = \phi \psi$ follows $\varphi = \psi \,.$

And $\phi$ is surjective, if and only if for any functions $\varphi , \psi \, : \, H \longrightarrow L$ to a set $L$ with $\varphi \phi = \psi \phi$ follows $\varphi = \psi \,.$

If you like you can show the equivalence of these definitions to the usual ones as an exercise. So injectivity is left cancellation and surjectivity right cancellation. For an isomorphisms we need, resp. have both. That's why I said in an earlier thread, that both directions are needed: $\phi \phi^{-1} = \operatorname{id}_H$ and $\phi^{-1} \phi = \operatorname{id}_G$.

#### Math_QED

Homework Helper
You can't talk about inverse functions if the function is not injective...

"Showing that inverse of an isomorphism is an isomorphism"

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