Exercise 28c.

A group $G$ is isomorphic to a group $G'$ if there exists a one-to-one correspondence $\phi \colon G \rightarrow G'$ such that $\phi(ab)=\phi(a)\phi(b)$, or in other words, such that $\phi$ preserves group products. Show that such a mapping $\phi$ (called an isomorphism) preserves identity elements and inverses.

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Proof

We need to show two things:

1. That $\phi$ preserves the identity element: $\phi(e_G)\phi(g)=\phi(g)=\phi(g)\phi(e_G)$
2. That $\phi$ preserves the inverse element: $\phi(g^{-1})\phi(g)= \phi(e_G)=\phi(g)\phi(g^{-1})$

where $e_G$ is the identity element of $G$.

Proof of 1

Let $G' \neq \varnothing$ and let $g' \in G'$. Since $\phi$ is a one-to-one correspondence (a bijection), there exists a $g \in G$ such that $g'=\phi(g)$. Now, since $\phi$ preserves the group product, we have the following $$g'=\phi(g)=\phi(e_Gg)=\phi(e_G)\phi(g)=\phi(e_G)g'$$ We have shown that $\phi(e_G)g'=g'$, and we get $g'\phi(e_G)=g'$ analogously, which means that $\phi$ preserves the identity element, which means that $\phi(e_G)$ is the identity element of the group $G'$. The proof of 1. is complete.

Proof of 2

Let $g \in G$. Since $G$ is a group, $g$ has an inverse $g^{-1} \in G$. Consider the following equalities, which utilize the fact that $\phi$ preserves the group product: $$\phi(g^{-1})\phi(g) = \phi(g^{-1}g) = \phi(e_G)$$ We got that $\phi(g^{-1})\phi(g)=\phi(e_G)$, and we can analogously show that $\phi(e_G)=\phi(g)\phi(g^{-1})$. This shows that $\phi(g^{-1})$ is the inverse of $\phi(g)$ (in the group product sense), that is, $\phi(g^{-1}) = \phi(g)^{-1}$.

We've proven both 1. and 2., the isomorphism $\phi$ indeed preserves the identity and inverse elements. The proof is complete.

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Additional thoughts