A subgroup having index $n$ containing no normal subgroup is isomorphic to $S(n)$ If $H$ be a subgroup of $G$ such that index of $H$ in $G$ is $n$, also $H$ does not contain any normal subgroups of $G$. Show that $H$ is isomorphic to a subgroup of $S(n)$.
$S(n)$ is the group of all permutations on a set of $n$ elements.
I don't understand where is the condition '$H$ does not contain any non trivial normal subgroup' useful?
I am thankful in advance to anyone who will make me understand.
Thank You.
 A: Write $G=t_1H\cup t_2H \cup \dots\cup t_nH$.
We define $\phi:G\rightarrow S_n$ by $$\phi(x)=\begin{pmatrix}t_1H&&t_2H&&\dots &&t_nH\\
xt_1H&&xt_2H&&\dots &&xt_nH\end{pmatrix}$$
It can be verified that $\phi$ is a homomorphism where the kernel of $\phi$ consists of the intersection of all the groups conjugate with $H$.
Let $K=\ker \phi$
Clearly, $K$ is a normal subgroup of $G$.
So $K$ is a normal subgroup of $H$ since $K=\cap_{g\in G}g^{-1}Hg\leq H$
Since $H$ does not contain any nontrivial normal subgroup, $K=\{1\}$, which implies that $\phi$ is injective.
We conclude that $\phi$ is an isomorphism from $G$ onto a subgroup of $S_n$
Extra explanation:
Let $x\in \ker \phi$
Then $xt_iH=t_iH$ which means that $x\in t_iHt_i^{-1}$ for $i=1,\dots,n$
Let $y\in G$
Then $y=t_ih$ for some $i$
Hence $yHy^{-1}=t_ihHh^{-1}t_i^{-1}=t_iHt_i^{-1}$.
Thus every conjugate of $H$ is in the form of $t_iHt_i^{-1}$.
Since $x\in \ker\phi$ lies in $t_iHt_i^{-1}$ for every $i=1,2,\dots,n$, we conclude that $x$ must lie in every conjugate of $H$.
