Lemma 8.15

Statement:

Let $G$ and $G'$ be groups and let $\phi:G \rightarrow G'$ be a one-to-one function such that $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in G$.Then $\phi[G]$ is a subgroup of $G'$ and $\phi$ provides an isomorphism of $G$ with $\phi[G]$.

Proof:

We show the conditions for a subgroup given in Theorem 5.14 are satisfied by $\phi[G]$.Let $x',y\prime \in \phi[G]$.Then there exist $x,y\in G$ such that $\phi(x)=x'$ and $\phi(y)=y'$.By hypothesis,$\phi(xy)=\phi(x)\phi(y)=x'y'$,showing that $x'y'\in\phi[G]$.We have shown that $\phi[G]$ is closed under the operation of $G'$.
Let $e'$ be the identity of $G'$.Then

(1)
\begin{align} e'\phi(e)=\phi(e)=\phi(ee)=\phi(e)\phi(e). \end{align}

Cancellation in $G'$ shows that $e'=\phi(e)$ so $e'\in\phi[G]$.
For $x'\in\phi[G]$ where $x'=\phi(x)$,we have

(2)
\begin{align} e'=\phi(e)=\phi(xx^{-1})=\phi(x)\phi(x^{-1})=x' \phi(x^{-1}), \end{align}

which shows that $x'^{-1}=\phi(x^{-1})\in\phi[G]$ .This completes the demonstration that $\phi[G]$ is a subgroup of $G'$.
That $\phi$ provides an isomorphism of $G$ with $\phi[G]$ now follows at once because $\phi$ provides a one-to-one map of $G$ onto $\phi[G]$ such that $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in G$.