Any subgroup of index $p$ in a $p$-group is normal.

Solution 1:

I think another approach in light of Don's answer can be:

Lemma: Suppose $G$ is a $p$-group and $H < G$, then $H\lneqq N_G(H)$.

Here we know that $[G:H]=p$ then $H$ is a proper subgroup of $G$. So the lemma tells us in this group we have $H$ as a proper subgroup of its normalizer in $G$. In fact our conditions make $N_G(H)$ be $G$ itself and this means that $H\vartriangleleft G$.

Solution 2:

You only need the following

Lemma:: If $\,G\,$ is a finite group and $\,p\,$ is the smallest prime diving $\,|G|\,$ , then any subgroup of index $\,p\,$ is normal in $\,G\,$ .

Proof (highlights): Let $\,N\leq G\,\,,\,\,[G:N]=m\,$ , and define an action of $\,G\,$ on the set of $\,X\,$ of left cosets of $\,N\,$ by $\,g\cdot(xN):=(gx)N\,$ :

1) Check the above indeed is a group action on that set

2) Check that the given action induces a homomorphism $\,\phi:G\to S_X\cong S_m\,$ with kernel

$$\ker\phi=\bigcap_{x\in G}N^x\,,\,\,\,N^x:=x^{-1}Nx $$

(the above kernel is also called the core of $\,N\,$)

3) Check that $\,\ker\phi\,$ is the greatest subgroup of $\,G\,$ normal in $\,G\,$ which is contained in $\,N\,$

4) Now apply the above to the case $\,m=p=\,$ the smallest prime dividing the order of the group.