TITLE: If $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m <n.$
QUESTION [12 upvotes]: Prove that if $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m <n.$

Since $G$ is of prime-power order I know $|Z(G)| \ne e$ so there is an $a\in Z(G)$ with order $p$ such that $p \mid |Z(G)|$. Now, the subgroup generated by is normal since it's a subgroup of the center. How can I get this normal subgroup to be less than $n$?

REPLY [5 votes]: Since you have a central element$~a$ of order$~p$, this is immediate by induction on$~n$. For $n=0$ there are no suitable $m$ so there is nothing to prove. For $n>0$ and $m=0$ take the subgroup $\{e\}$. Otherwise by induction there is a subgroup of order $p^{m-1}$ in $G/\langle a\rangle$ and its inverse image in$~G$ has order$~p^m$.