Theorem 2.G (p. 62, chapter 2):If $G$ is a group, $H$ a subgroup of $G$, and $S$ is the set of all right cosets of $H$ in $G$, then there is a homomorphism $\theta$ of $G$ into $A(S)$, and the kernel of $\theta$ is the largest normal subgroup of $G$ which is contained in $H$.
(a few words about notation: Herstein uses $A(S)$ to stand for the group of all bijections on $S$...if $|S| = n$, then $A(S)$ is isomorphic to $S_n$. Herstein also writes his mappings on the RIGHT, as in $(x)\sigma$ instead of $\sigma(x)$, so that composition and multiplication are "in the same order", instead of reversed. For this reason, he uses right cosets and right-multiplication instead of the left cosets (and left-multiplication) one often sees used in other texts. He also denotes the index of $H$ in $G$ by $i(H)$ , instead of $[G
]$ and denotes $|G|$ by $o(G)$).
Proof: Let $G$ be a group, $H$ a subgroup of $G$. Let $S$ be the set whose elements are right cosets of $H$ in $G$. That is, $S = \{Hg: g \in G\}$. $S$ need not be a group itself, in fact, it would be a group only if $H$ were a normal subgroup of $G$. However, we can make our group $G$ act on $S$ in the following natural way: for $g \in G$ let $t_g:S \to S$ be defined by: $(Hx)t_g = Hxg$. Emulating the proof of Theorem 2.f we can easily prove:
(1) $t_g \in A(S)$ for every $g \in G$
(2) $t_{gh} = t_gt_h$.
Thus the mapping $\theta: G \to A(S)$ defined by $\theta(g) = t_g$ is a homomorphism of $G$ into $A(S)$. Can one always say that $\theta$ is an isomorphism? Suppose that $K$ is the kernel of $\theta$. If $g_0 \in K$, then $\theta(g_0) = t_{g_0}$ is the identity map on $S$, so that for every $X \in S, Xt_{g_0} = X$. Since every element of $S$ is a right coset of $H$ in $G$, we must have that $Hat_{g_0} = Ha$ for every $a \in G$, and using the definition of $t_{g_0}$, namely, $Hat_{g_0} = Hag_0$, we arrive at the identity $Hag_0 = Ha$ for every $a \in G$. On the other hand if $b \in G$ is such that $Hxb = Hx$ for every $x \in G$, retracing our argument we could show that $b \in K$. Thus $K = \{b \in G|Hxb = Hx$ all $x \in G\}$. We claim that from this characterization of $K,\ K$ must be the largest normal subgroup of $G$ which is contained in $H$. We first explain the use of the word largest; by this we mean if $N$ is a normal subgroup of $G$ which is contained in $H$, then $N$ must be contained in $K$. We wish to show this is the case. That $K$ is a normal subgroup of $G$ follows from the fact that it is the kernel of a homomorphism of $G$. Now we assert that $K \subset H$, for if $b \in K, Hab = Ha$ for every $a \in G$, so in particular, $Hb = Heb = He = H$, whence $b \in H$. Finally, if $N$ is a normal subgroup of $G$ which is contained in $H$, if $n \in N,\ a \in G$, then $ana^{-1} \in N \subset H$, so that $Hana^{-1} = H$; thus $Han = Ha$ for all $a \in G$. Therefore, $n \in K$ by our characterization of $K$.
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Remarks following the proof:
The case $H = (e)$ just yields Cayley's Theorem (Theorem 2.f). If $H$ should happen to have no normal subgroup of $G$, other than $(e)$ in it, then $\theta$ must be an isomorphism of $G$ into $A(S)$...(some text omitted)...
We examine these remarks a little more closely. Suppose that $G$ has a subgroup $H$ whose index $i(H)$ (that is, the number of right cosets of $H$ in $G$) satisfies $i(H)! < o(G)$. Let $S$ be the set of all right cosets of $H$ in $G$. The mapping, $\theta$, of Theorem 2.g cannot be an isomorphism, for if it were, $\theta(G)$ would have $o(G)$ elements and yet would be a subgroup of $A(S)$ which has $i(H)! < o(G)$ elements. Therefore, the kernel of $\theta$ must be larger than $(e)$; this kernel being the largest normal subgroup of $G$ which is contained in $H$, we can conclude that $H$ contains a nontrivial normal subgroup of $G$.
However, the above argument has implications even when $i(H)!$ is not less than $o(G)$. If $o(G)$ does not divide $i(H)!$ then by invoking Lagrange's theorem we know that $A(S)$ can have no subgroup of order $o(G)$, hence no subgroup isomorphic to $G$. However $A(S)$ does contain $\theta(G)$, whence $\theta(G)$ cannot be isomorphic to $G$, that is, $\theta$ cannot be an isomorphism. But then, as above, $H$ must contain a nontrivial normal subgroup of $G$. We summarize this as:
Lemma 2.21 If $G$ is a finite group, and $H \neq G$ is a subgroup of $G$ such that $o(G) \not\mid i(H)!$, then $H$ must contain a nontrivial normal subgroup of $G$. In particular, $G$ cannot be simple.
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