injection can be extended to isomorphism

Theorem.  If $f$ is an injection from a set $S$ into a group $G$, then there exist a group $H$ containing $S$ and a group isomorphism  $\phi :H\to G$  such that  ${\phi |}_S=f$.

Proof.  Let $M$ be a set such that  $\mathrm{card}(M)\geqq \mathrm{card}(G)$.  Because  $\mathrm{card}(f(S))=\mathrm{card}(S)$, we have $\mathrm{card}(M\setminus S)\geqq \mathrm{card}(G\setminus f(S))$,  and therefore there exists an injection

$$\psi :G\setminus f(S)\to M\setminus S$$

(provided that  $G\setminus f(S)\ne \mathrm{\varnothing }$;  otherwise the mapping  $f:S\to G$ would be a bijection).  Define

$$H:=S\cup \psi (G\setminus f(S)),$$

$\phi (h):=\{\begin{array}{cc}f(h)\hspace{1em}\hspace{1em}\text{for}h\in S,\hfill & \\ {\psi }^{-1}(h)\hspace{1em}\text{for}h\in H\setminus S.\hfill & \end{array}$

Then apparently,  $\phi :H\to G$  is a bijection and  ${\phi |}_S=f$.  Moreover, define the binary operation “$*$” of the set $H$ by

$h_1\ast h_2:={\phi }^{-1}(\phi (h_1)\cdot \phi (h_2)).$(1)

We see first that

$(h_1\ast h_2)\ast h_3$$\mathrm{\hspace{0.33em}}={\phi }^{-1}\left(\phi \left({\phi }^{-1}(\phi (h_1)\cdot \phi (h_2))\right)\cdot \phi (h_3)\right)$

$\mathrm{\hspace{0.33em}}={\phi }^{-1}((\phi (h_1)\cdot \phi (h_2))\cdot \phi (h_3))$

$\mathrm{\hspace{0.33em}}={\phi }^{-1}(\phi (h_1)\cdot (\phi (h_2)\cdot \phi (h_3)))$

$\mathrm{\hspace{0.33em}}={\phi }^{-1}\left(\phi (h_1)\cdot \phi \left({\phi }^{-1}(\phi (h_2)\cdot \phi (h_3))\right)\right)$

$\mathrm{\hspace{0.33em}}=h_1\ast (h_2\ast h_3).$

Secondly,

$$h\ast {\phi }^{-1}(e)={\phi }^{-1}\left(\phi (h)\cdot \phi ({\phi }^{-1}(e))\right)={\phi }^{-1}(\phi (h))=h,$$

whence ${\phi }^{-1}(e)$ is the right identity element of $H$.  Then,

$$h\ast {\phi }^{-1}\left({(\phi (h))}^{-1}\right)={\phi }^{-1}\left(\phi (h)\cdot \phi \left({\phi }^{-1}(\phi {(h)}^{-1})\right)\right)={\phi }^{-1}(e),$$

and accordingly ${\phi }^{-1}\left({(\phi (h))}^{-1}\right)$ is the right inverse of $h$ in $H$.  Consequently,$(H,\ast )$ is a group.  The equation (1) implies that

$$\phi (h_1\ast h_2)=\phi (h_1)\cdot \phi (h_2),$$

whence $\phi$ is an isomorphism from $H$ onto $G$.  Q.E.D.