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According to this image:

enter image description here

I would like to know why isomorphism implies that if $AB=C$ then $A'B'=C'$

To do this, I have to prove that the isomorphism, call it f, has to satisfy: $f(AB) = f(A)f(B)$, but how do I prove this?

cazanova
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    IMO, that "in other words" is misleading: what follows is not the same as stated till that point (bijectivity), but just an additional, key feature of "isomorphism" definition, namely operation preservation. –  Apr 09 '21 at 06:27
  • I have looked for group isomorphism in https://math.stackexchange.com/questions/147632/isomorphisms-preserve-structure-operation-or-order

    But how do I prove that an isomorphism satisfies these conditions? Is it just defined like that?

     – cazanova Apr 09 '21 at 06:32
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    Yes, the isomorphism is defined like that. You can define that two groups are isomorphic based on a diffrent definition, and come up to the lemma that two groups are isomorphic iff there is an operation-preserving bijection between them. But this is another story. –  Apr 09 '21 at 06:36

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As mentioned in the comments, the definition of a group isomorphism in the quoted text is incorrect. The standard definition is as follows:

Let $G$ and $H$ be groups.

  • A function $\varphi: G \to H$ is a group homomorphism if for all $x , y \in G$, $\varphi(xy) = \varphi(x)\varphi(y)$.
  • A function $\varphi : G \to H$ is a group isomorphism if it is a group homomorphism and a bijection (equivalently $\varphi$ is a bijection and there exists a group homomorphism $\psi : H \to G$ such that $\varphi \circ \psi = \operatorname{id}_H$ and $\psi \circ \varphi = \operatorname{id}_G$).

Therefore the desired $\varphi(gh) = \varphi(g)\varphi(h)$ is simply part of the definition.

Nate Gallup
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