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

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 $n$?

• You can induct on $n$ and use the fact that $\left |G/Z(G)\right | < |G|$ since $Z(G)\neq \{e\}$ – Prahlad Vaidyanathan Nov 14 '13 at 17:14
• This is a theorem in every Group theory book which contains a chapter about sylow theorems. – mrs Nov 14 '13 at 17:16
• It's probably “a subgroup of order $p^m$”. – egreg Nov 14 '13 at 17:16
• @B.S. You don't need the Sylow theorems to prove this. Merely the class equation (which implies that $Z(G) \neq \{e\}$ in this case) – Prahlad Vaidyanathan Nov 14 '13 at 17:18

## 2 Answers

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.

• Sylow is often only formulated for maximal prime powers, since the result for smaller powers is much easier once the one for maximal powers has been established. Cauchy's theorem and the classification of finite abelian groups are both way overkill compared to what is necessary for this. All you need is that the center is non-trivial and induction. – Tobias Kildetoft Nov 14 '13 at 20:32
• True, while I realized this was overkill given the order of $|G|$, it is a useful exercise to be able to go through this proof. It was even placed on one of my prelims at Cornell! But yes, for the OP you really need only continue to factor our subgroups of order $p$ using induction. – mathematics2x2life Nov 14 '13 at 20:36
• @user104235 Not a problem. This is a way to do it and I'd encourage you to try it! But also note that there are shorter ways to do this that would be also good to know as well! But knowing this proof of Sylow's First Theorem is also important. – mathematics2x2life Nov 15 '13 at 2:31
• After using induction hypothesis in step 5, how do I come back from $G/\langle a\rangle$ to G? – emmy Dec 24 '16 at 10:27
• Also why go to $Z(G)$ to find $x$ of order p? You can use Cauchy's theorem to get this x in $G$ itself ? Is there any need for $\langle x \rangle$ to be normal ? – emmy Dec 24 '16 at 10:32

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$~p^m$.

• The inverse image, is it well defined ? Each element in $G/\langle a \rangle$ maps to more than one element of $G$. – emmy Dec 24 '16 at 10:24
• @emmy The inverse image of a subgroup $S$ of $H$ under a group morphism $f:G\to H$ is $S'=\{\,g\in G\mid f(g)\in S\,\}$. This is always well defined and always a subgroup of $G$. Moreover $S\cong S'/\ker(f)$ by the first isomorphism theorem, and in particular $|S'|=|\ker(f)|.|S|$. This is applied in the answer for $f:G\to G/\langle a\rangle$ the canonical projection, and $S$ the (unnamed) subgroup of order $p^{m-1}$ of $G/\langle a\rangle$. – Marc van Leeuwen Dec 24 '16 at 11:24