injection can be extended to isomorphism

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$\varphi\!:H\to G$  such that  $\varphi|_{S}=f$.

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

 $\psi\!:G\!\smallsetminus\!f(S)\to M\!\smallsetminus\!S$

(provided that  $G\!\smallsetminus\!f(S)\neq\varnothing$;  otherwise the mapping$f\!:S\to G$ would be a bijection).  Define

 $H\;:=\;S\cup\psi(G\!\smallsetminus\!f(S)),$
 $\displaystyle\varphi(h)\;:=\;\begin{cases}f(h)\qquad\mbox{for}\;\;h\in S,\\ \psi^{-1}(h)\quad\mbox{for}\;\,h\in H\!\smallsetminus\!S.\end{cases}$

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

 $\displaystyle h_{1}\ast h_{2}\;:=\;\varphi^{-1}(\varphi(h_{1})\!\cdot\!\varphi% (h_{2})).$ (1)

We see first that

 $\displaystyle(h_{1}\ast h_{2})\ast h_{3}$ $\displaystyle\;=\;\varphi^{-1}\!\left(\varphi\left(\varphi^{-1}(\varphi(h_{1})% \!\cdot\!\varphi(h_{2}))\right)\!\cdot\!\varphi(h_{3})\right)$ $\displaystyle\;=\;\varphi^{-1}((\varphi(h_{1})\!\cdot\!\varphi(h_{2}))\!\cdot% \!\varphi(h_{3}))$ $\displaystyle\;=\;\varphi^{-1}(\varphi(h_{1})\!\cdot\!(\varphi(h_{2})\!\cdot\!% \varphi(h_{3})))$ $\displaystyle\;=\;\varphi^{-1}\!\left(\varphi(h_{1})\!\cdot\!\varphi\left(% \varphi^{-1}(\varphi(h_{2})\!\cdot\!\varphi(h_{3}))\right)\right)$ $\displaystyle\;=\;h_{1}\ast(h_{2}\ast h_{3}).$

Secondly,

 $h\ast\varphi^{-1}(e)\;=\;\varphi^{-1}\!\left(\varphi(h)\!\cdot\!\varphi(% \varphi^{-1}(e))\right)\;=\;\varphi^{-1}(\varphi(h))\;=\;h,$

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

 $h\ast\varphi^{-1}\!\left((\varphi(h))^{-1}\right)\;=\;\varphi^{-1}\left(% \varphi(h)\!\cdot\!\varphi\left(\varphi^{-1}(\varphi(h)^{-1})\right)\right)\;=% \;\varphi^{-1}(e),$

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

 $\varphi(h_{1}\ast h_{2})\;=\;\varphi(h_{1})\!\cdot\!\varphi(h_{2}),$

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

Title injection can be extended to isomorphism InjectionCanBeExtendedToIsomorphism 2013-03-22 18:56:50 2013-03-22 18:56:50 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 20A05 msc 03E20 Restriction Cardinality