Could anyone give me a hint on how to solve the following problem?
Prove that if $p$ is a prime and $G$ is a group of order $p^{\alpha}$ for some $\alpha \in \mathbb{Z}^{+}$, then every subroup of index $p$ is normal in $G$.
I don't know where to start to solve this problem
We can use an inductive argument, since the claim is trivially true for $\;G\;$ with $\;|G|=p\;,\;\;or\;\;p^2\;$ Suppose $\;H\le G\;$ is a subgroup of index $\;p\;$ (and thus maximal as proper subgroup), so we have two possibilities:
$$H\lneq N_G(H)\implies N_G(H)=G\iff H\lhd G\;\;,\;\;\text{and we're done.}$$
Or else (put $\;Z:=Z(G)=$ the center of $\;G\;$) :
$$H=N_G(H)\implies\;Z\le N_G(H)=H,\;\text{Inductively}\;,\;\;H/Z\lneq N_{G/Z}(H/Z)=G/Z\;,\;\;\text{since}$$
$$[(G/Z):(H/Z)]=[G:H]=p\;\implies\;\exists\,x\in G\;\;s.t.\;\;xZ\in N_{G/Z}(H/Z)\setminus H/Z$$
and this means $\;x\in G\setminus H\;$ and fulfills
$$(xZ)(hZ)(xZ)^{-1}=(xhx^{-1})Z\in H/Z\iff xhx^{-1}\in H\implies x\in H\implies\;\text{contradiction}$$
Google "normalizer condiction". For finite groups, this condition is equivalent to the group being nilpotent, which any finite $\;p\,-$ group is.
