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**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

Cancellation in $G'$ shows that $e'=\phi(e)$ so $e'\in\phi[G]$.

For $x'\in\phi[G]$ where $x'=\phi(x)$,we have

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 $x,y\in G$.