to get the group of order $\displaystyle p^{n-1}$ I will use induction.

Base case n=1 $\displaystyle |G|=p$

case n=2 $\displaystyle |G|=p^2$ would be covered by the above argument

Asume true for n

$\displaystyle |G|=p^n$ has a normal subgroup H of order $\displaystyle |H|=p^{n-1}$

Show n+1

$\displaystyle |G|=p^{n+1}$ so by the first argument this has a normal subgroup of order P call it N.

So then $\displaystyle |G/N|=p^n$ by the induction hypothsis this has a normal subgroup H of order $\displaystyle |H|=p^{n-1}$ so

$\displaystyle H/N \underline\vartriangleleft G/N $ Now by the 4th isomorphism theorem(lattice) $\displaystyle H \underline\vartriangleleft G$ and $\displaystyle |H|=p^n$ and this completes the induction.