Subgroups of order $p$ and $p^{n-1}$ in a group of order $p^n$. I have a group $G$ of order $p^n$ for $n \ge 1$ and $p$ a prime.  I am looking for two specific subgroups within $G$: one of order $p$ and one of order $p^{n-1}$. I don't think I would use the Sylow theorems here because those seem to apply to groups with a "messier" order than simply $p^n$. Would Cauchy's Theorem allow me to generate the two requisite subgroups? I could use it to find an element of order $p$ and an element of order $p^{n-1}$ and then consider the cyclic subgroups generated by these two elements? 


 Let $P$ act on itself by conjugation. $1$ appears in an orbit of size $1$, and everything else appears in an orbit of size $p^k$ for some $k$.  Since the sum of the orbit sizes is equal to $|P|$, which is congruent to $0\mod{p}$, that means there has to be at least one more orbit of size $1$.  Orbits of size $1$ under conjugation contain elements which commute with everything in the group; they compose $Z(P)$.  Now suppose inductively that $\exists S\unlhd P$ with $|S|=p^k$.  Then by the above lemma and Cauchy's theorem $P/S$ has a central subgroup $\overline{Q}$ of order $p$.  Lifting $\overline{Q}$ back to $P$, we obtain a normal subgroup of order $p^{k+1}$.

