Subgroup with index equal to smallest prime factor normal. How can I prove this?

Question

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.

Answer 1

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 $H$.

Answer 2

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 $G$.

Answer 3

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.

Answer 4

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 $H\trianglelefteq G$.