Exercise 7.4.4

Proof.

(a)

Let $${\phi }_g:\{\begin{array}{ccc}\hfill G\hfill & \hfill \to \hfill & \hfill G\hfill \\ \hfill h\hfill & \hfill \mapsto \hfill & \hfill \mathit{gh}\hfill \end{array}$$

$\text{∙}$ ${\phi }_g$ is injective: let $h,k\in G$.

If ${\phi }_g(h)={\phi }_g(k)$, then $\mathit{gh}=\mathit{gk}$, therefore $g^{-1}\mathit{gh}=g^{-1}\mathit{gk}$, $h=k$.

$$\forall <!--\; FUNCTION\; APPLICATION\; -->h\in G,\forall <!--\; FUNCTION\; APPLICATION\; -->k\in G,{\phi }_g(h)={\phi }_g(k)\Rightarrow h=k.$$

$\text{∙}$ ${\phi }_g$ is surjective: let $k$ be any element in $G$.

Put $h=g^{-1}k$. Then ${\phi }_g(h)=g(g^{-1}k)=(g{g}^{-1})k=\mathit{ek}=k$.

$$\forall <!--\; FUNCTION\; APPLICATION\; -->k\in G,\exists <!--\; FUNCTION\; APPLICATION\; -->h\in G,{\phi }_g(h)=k.$$

(b)

A row of the Cayley table of $G$ corresponding to the element $g\in G$ is the list of the ${\phi }_g(g_i)=g{g}_i$, where $g_i$ traces the list of the elements of $G$ in an arbitrary fixed order. Since ${\phi }_g$ is bijective, we find all the elements of $G$ once and only once. This defines a permutation of $G$.

(c)

Write $S(G)$ the group of bijections of $G$ in $G$, and $S_n$ the group of bijections of $\text{[[}1,n\text{]]}$ in $\text{[[}1,n\text{]]}$ (where $\text{[[}1,n\text{]]}=\{1,2,\cdots ,n\}$).

The map $\phi :G\to S(G),g\mapsto {\phi }_g=\phi (g)$ is an injective group homomorphism.

Indeed, for all $g,h,k\in G$,

$(\phi (g)\circ \phi (h))(k)={\phi }_g({\phi }_h(k))=g(\mathit{hk})=(\mathit{gh})k=\phi (\mathit{gh})(k)$, thus

$$\phi (g)\circ \phi (h)=\phi (\mathit{gh}).$$

If $\phi (g)=1_G$, $e=\phi (g)(e)=\mathit{ge}=g$, therefore $g=e$: $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{e\}$.

Moreover, if $f:\text{[[}1,n\text{]]}\to G,i\mapsto g_i$ is the bijection representing the chosen numbering of $G$, we can associate to it the isomorphism

$$\psi :\{\begin{array}{ccc}\hfill S(G)\hfill & \hfill \to \hfill & \hfill S_n\hfill \\ \hfill u\hfill & \hfill \mapsto \hfill & \hfill f^{-1}\circ u\circ f\hfill \end{array}$$

where $\psi (u)=f^{-1}\circ u\circ f$ is indeed a permutation of $\text{[[}1,n\text{]]}$.

If $u,v\in S(G)$, $\psi (u)\circ \psi (v)=f^{-1}\circ u\circ f\circ f^{-1}\circ v\circ f=f^{-1}\circ (u\circ v)\circ f=\psi (u\circ v)$, so $\psi$ is a group homomorphism.

If $\psi (u)=e$, then $f^{-1}\circ u\circ f=e$, thus $u=f\circ f^{-1}=e$. Therefore $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )=\{e\}$.

Let $\sigma$ any permutation in $S_n$. Put $u=f\circ \sigma \circ f^{-1}$.

Then $\psi (u)=f^{-1}\circ f\circ \sigma \circ f^{-1}\circ f=\sigma$, thus $\psi$ is surjective. $\psi$ is a group isomorphism (depending of the chosen numbering).

Thus $\chi =\psi \circ \phi :G\to S_n$ is an injective group homomorphism.

For each $g_i\in G$, we associate to it ${\sigma }_i=\chi (g_i)$.

Let $k\in [1,n]$ defined by $g_i{g}_j=g_k$, which is equivalent to $k=f^{-1}(g_i{g}_j)$.

$$\begin{array}{llll}\hfill g_i{g}_j& ={\phi }_{g_i}(g_j)& \hfill & \\ \hfill & =({\phi }_{g_i}\circ f)(j),& \hfill & \end{array}$$

therefore

$$\begin{array}{llll}\hfill k& =f^{-1}(g_i{g}_j)& \hfill & \\ \hfill & =(f^{-1}\circ {\phi }_{g_i}\circ f)(j)& \hfill & \\ \hfill & =[\psi ({\phi }_{g_i})](j)& \hfill & \\ \hfill & =[(\psi \circ \phi )(g_i)](j)& \hfill & \\ \hfill & ={\sigma }_i(j).& \hfill & \end{array}$$

IIf ${\sigma }_I=\chi (g_i)$, we have so proved that for all $i,j\in [1,n]$,

$$g_i{g}_j=g_{{\sigma }_i(j)}.$$