Exercise 14.3.9

Proof. Recall the construction of $G\simeq H\subset S_n$ (see ex. 7.4.4, 7.4.6).

Let $f:\{1,\dots ,n\}\to G$ be a bijection, and write $g_i=f(i)$, so that $G=\{g_1,\dots ,g_n\}$.

For each $i\in \{1,\dots ,n\}$ there is some permutation ${\sigma }_i$ such that

Consider the maps

where $G^{\prime }=\{{\varphi }_g\mid g\in G\}\subset S(G),H\subset S_n$, and $G\simeq H$. Here $f,\varphi ,\psi$ are bijective, and $\varphi ,\psi$ are group isomorphisms (Ex .7.4.4).

Note that, as already seen in ex. 7.4.4, for all $j\in \{1,\dots ,n\}$,

thus

and $H=\{{\sigma }_1,\dots ,{\sigma }_n\}$.

Write $\delta =\psi \circ \varphi \circ f$, so that $\delta (i)=(\psi \circ \varphi \circ f)(i)=f^{-1}\circ {\varphi }_{g_i}\circ f,i=1,\dots ,n.$

We define $\gamma ={\delta }^{-1}$, thus for all $i\in \{1,\dots ,n\}$,

Here $T=\{1,\dots ,n\}$, since $H\subset S(\{1,\dots ,n\})$. Then $\gamma :G\to T$ induces an isomorphism $\hat{\gamma }:S(G)\to S(T)=S_n$ defined by

Now we compute $\hat{\gamma }({\phi }_h)$, where $h\in H$ is a permutation, and ${\phi }_h(k)=\mathit{hk}=h\circ k,k\in H$. Since $h\in H$, there is some $i\in \{1,\dots ,n\}$ such that $h=f^{-1}\circ {\varphi }_{g_i}\circ f={\sigma }_i$.

For all $j\in \{1,\dots ,n\}$,

Therefore $\hat{\gamma }({\phi }_h)={\sigma }_i=h\in H$.

We have proved that the isomorphism $\hat{\gamma }$ takes $\{{\phi }_h\mid h\in H\}$ in $H$, and since these two subgroups have same order $n$,

The definition of a regular subgroup of $S(T)$ is satisfied, so $H$ is a regular subgroup of $S_n$. □