Originally Posted by

**tangibleLime** Hm, I think I understand what you are saying. So let me try again.

It is given that $\displaystyle \phi : G \rightarrow G'$ is an isomorphism. Need to prove that $\displaystyle \phi^{-1}: G' \rightarrow G$ is an isomorphism.

Since $\displaystyle \phi$ defines an isomorphism, it satisfies the homomorphism property:

$\displaystyle \phi(xy) = \phi(x)\phi(y) \hspace{4mm} \forall x,y \in G$

And $\displaystyle \phi$ maps values from G to G',

$\displaystyle \phi(x) = x'$, where $\displaystyle x \in G$ and $\displaystyle x' \in G'$.

Since $\displaystyle \phi$ is bijection, and therefore onto,

$\displaystyle \phi^{-1}(\phi(x)) = x$

$\displaystyle \Rightarrow \phi^{-1}(x') = x$

Using this fact with the homomorphism property,

$\displaystyle \phi^{-1}(\phi(xy)) = \phi^{-1}(\phi(x))\phi^{-1}(\phi(y))$

$\displaystyle \Rightarrow \phi^{-1}(x'y') = \phi^{-1}(x')\phi^{-1}(y')$

$\displaystyle \Rightarrow xy = xy$

And so $\displaystyle \phi^{-1}$ satisfies the homomorphism property, and is an isomorphism since it has an inverse $\displaystyle \phi$.

Am I close?