Isomorphism and multiplication table

Question

According to this image:

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?

Answer 1

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.