Subgroup with index equal to smallest prime factor normal. How can I prove this? Let $G$ be a group of order $n>1$ and $p$ the smallest prime factor of $n$. Suppose, $H$ is a subgroup of $G$ and $[G:H]=p$.

How can I prove that $H$ is normal ?
According to Lagrange, this means that every subgroup $H$ of order $\frac{n}{p}$ is normal, right ?

One property of a normal subgroup $H$ is that for every $a\in G$ and $b\in H$, we have $a^{-1}ba\in H$, but I have no idea how I can show this for the subgroup above.
 A: Consider the action $G$ on the cosets of $H$, giving rise to a map $G \to S_p$. Clearly the kernel is contained in $H$, hence the index of the kernel (which is the size of the image)  is divisible by $p$.
Now consider the size of the image again. It divides $p!$ and $|G|$, hence it divides $p$ by the assumption.
The two arguments above show that the index of the kernel is $p$, hence the kernel is $H$.
A: Hint. Let $G$ be a finite group and $H$ be a subgroup of $G$. Let define the following group homomorphism: $$h:\left\{\begin{array}{ccc}G&\rightarrow&\mathfrak{S}(G/H)\\g&\mapsto&h(g):aH\mapsto gaH\end{array}\right..$$
Notice that:


*

*$[G:H]!$ is divided by $[G:\textrm{ker}(h)]$.

*$\textrm{ker}(h)\subseteq H$.


Which implies that $([G:H]-1)!$ is divided by $[H:\textrm{ker}(h)]$.
If you assume that the index of $H$ in $G$ is the smallest prime factor of the order of $G$, you will manage to prove that $H=\textrm{ker}(h)$ and hence $H$ is normal in $G$.
A: Let $(G/H)_l$ $\quad$ the set of left cosets.
Let f:G $\rightarrow$ $\Sigma(G/H)$,$\quad$$f(g)=f_g$$\quad$where $f_g(xH)=gxH$ $\qquad$ and$\quad$
$\Sigma(G/H)_l$$\quad$=symmetries of$ (G/H)_l$$\quad$.
We have that f is morphism of groups and
Ker(f) is subgroup in H.By isomorphism  theorem         |G/Ker(f)|=|Im(f)|. Obviously |$\Sigma(G/H)_
l$$\quad$|=p!$\quad$.
So |G/Ker(f)| |p!.
It is easy to see that |G/Ker(f)|=p.
But Ker(f)$\le$H$\le$G.
It results that H=Ker(f).
Hence H is a normal subgroup.
A: because $[G:H]=p$ then according to the theorem of poincaré, if $N=\cap_{x\in G} xHx^{-1}$ then $[G:N]$ divise $p!$, because the smallest prime factor of $n$ is $p$, then $[G:N]=1$ or $p$, but $N\subset H$, so $[G:N]=[G:H][H:N]=p[H:N]$.
So $[G:N]$ is not equal to $1$ whence $[G:N]=p$. So $[H:N]=1$, which means $N=H$, finally $H\trianglelefteq G$.
