If $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m <n.$

Question

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 $n$?

Answer 1

First, I believe you meant that if $|G|=p^n$ then $G$ has a subgroup of order $p^m$ for all $0\leq m <n$.

This follows directly from Sylow's 1st Theorem which gives the existence of subgroups of prime-power order.

Sylow's First Theorem: Let $G$ be a finite group and $p$ be a prime. If $p^k$ divides $|G|$, then $G$ has at least one subgroup of order $p^k$.

So either you use this theorem directly along with the order of $G$ or you prove Sylow's Theorem and then apply it-depending on what your assignment was. If you have to prove it, I'll give some hints:

1. Use induction. What if $|G|=1$, i.e. $n=0$? Is the result true?
2. Assume that the result holds for any group with order less than $G$. If $G$ were to have a proper subgroup $H$ such that $p^k$ divides $|H|$, then use the induction hypothesis. What happens? Are you done?
3. After ($2$) you can assume that $p^k$ can't divide the order of any subgroup $H$ of $G$. Then use the class equation $$ |G|=|Z(G)|+\sum_{a} |G:C(a)| $$
   where the right sum is over representatives for the conjugacy class for $a \notin Z(G)$.
4. As $p^k$ divides $|G|=|G:C(a)|\,\,|C(a)|$, does $p$ divide $|Z(G)|$? Then use the Fundamental Theorem of abelian groups to find a subgroup of $Z(G)$ of order $p$ generated by say $x$.
5. Finally, look at $|G/\langle x\rangle|$. Is there a power of $p$ which divides the order of $|G/\langle x\rangle|$? If so, what does the induction hypothesis then imply? Does this complete the proof?

Though this is a longer way. There are faster ways using factor groups and Cauchy's Theorem. However, I don't know what you know/don't know and what you can use to complete the problem. The above proof alone is sufficient.

Answer 2

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$~p^m$.