Proving f inverse is homomorphic

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The discussion centers on proving that the inverse function f^(-1) of an isomorphism f: S -> S' is homomorphic. The user outlines the requirement to demonstrate that f^(-1)(x' *' y') equals f^(-1)(x') * f^(-1)(y'). They correctly identify that since f is bijective, f^(-1)(x') and f^(-1)(y') correspond to elements x and y in S. However, the user expresses uncertainty about simplifying f^(-1)(x' *' y').

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DPMachine
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I'm trying to show:

If f: S -> S' is an isomorphism of <S, *> with <S', *'>, then f^(-1) is homomorphic.

My take:

So I have to show that f^(-1)(x' *' y') = f^(-1)(x') * f^(-1)(y').

Since f is bijective (onto, more precisely) I know that f^(-1)(x') = x and f^(-1)(y') = y. So f^(-1)(x') * f^(-1)(y') = xy.

How can I simplify f^(-1)(x' *' y') though?

Thanks.
 
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EDIT: Nevermind, that's not right.

Sorry about not using latex, by the way. I was writing from a mobile device.
 

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