Theorem 8.16

Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. By Lemma 8.15, we need only to define a one-to-one function $\phi : G \rightarrow S_G$ such that $\phi (xy) = \phi (x) \phi (y)$ for all $x, y \in G$. For $x \in G$, let $\lambda _x : G \rightarrow G$ be defined by $\lambda _x (g) = xg$ for all $g \in G$. (We think of $\lambda _x$ as performing left multiplication by $x$.) The equation $\lambda _x (x^{-1} c) = x(x^{-1} c) =c$ for all $c \in G$ shows that $\lambda _x$ maps $G$ onto $G$. If $\lambda _x (a) = \lambda _x (b)$, then $xa = xb$ so $a = b$ by cancellation. Thus $\lambda _x$ is also one-to-one, and is a permutation of $G$. We now define $\phi : G \rightarrow S_G$ by defining $\phi (x) = \lambda _x$ for all $x \in G$.
To show that $\phi$ is one-to-one, suppose that $\phi (x) = \phi (y)$. Then $\lambda _x = \lambda _y$ as functions mapping $G$ into $G$. In particular $\lambda _x (e) = \lambda _y (e)$, so $xe = ye$ and $x=y$. Thus $\phi$ is one-to-one. It only remains to show that $\phi (xy) = \phi (x) \phi (y)$, that is, that $\lambda _{xy} = \lambda _x \lambda _y$. Now for any $g \in G$, we have $\lambda _{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda _x \lambda _y )(g) = \lambda _x (\lambda _y(g)) = \lambda _x (yg) =x(yg)$. Thus by associativity, $\lambda _{xy} = \lambda _x \lambda _y$. $\blacksquare$