1

I am reading Aluffi chapter 0. This is an exercise in it. Since the book hasn't stated Sylow theorem or Cauchy theorem yet, I don't want to use it in my solution, though I am stuck.

Let $p$ be a prime number, and let $G$ be a $p$-group : $|G| = p^r$. Prove that $G$ contains a normal subgroup of order $p^k$ for every $k \leq r$.

I know that the center isn't trivial by the class formula.

Shaun
  • 47,747
joe blacksmith
  • 431

1 Answers1

2

If $G$ is abelian, then this result follows from the theorem:

Theorem. Let $G$ be a finite abelian group of order $n$. Then $G$ has a subgroup of order $d$ for every divisor $d$ of $n$.

You can see the proofs from this link. The OP used Cauchy's theorem, but the proof of Cauchy's theorem for abelian groups is much easier than for non-abelian ones. Here I give a proof for that

Proof for Cauchy's theorem on Abelian Groups. Here we can prove it by induction on $n$. The case for $n=1$ is trivial. Now we let $a\in G$ be an element of order $k$.

  1. If $p\mid k$, say $k=p\ell$, then $a^\ell$ has order $p$.

  2. Suppose $p\nmid k$. Since $G$ is abelian, we have $N:=\langle a\rangle\unlhd G$. Now we consider $G/N$, which has order $n/k$. Since $p\mid n$ but $p\nmid k$, we must have $p\mid(n/k)$. By induction hypothesis, there is an element $bN\in G/N$ of order $p$. Clearly, the order of $b$ is a multiple of $p$, so everything reduces to the first case. $\square$

As for the followings, you can check that link.

Thus, we assume that $G$ is non-abelian and prove it by induction on $r$. The case for $r=1$ is trivial as $G$ is cyclic in this case.

As for the inductive step, the center $Z(G)$ of $G$ has size $p^\ell$ where $1\leq\ell<r$. We then consider two cases:

  1. If $k\leq\ell$, then we can find a subgroup $N$ of $Z(G)$ of order $p^k$ as $Z(G)$ is abelian. We can show that $N$ is a normal subgroup of $G$. (Hint. $N$ is contained in the center of $G$)

  2. If $k>\ell$, we consider the quotient group $G/Z(G)$, which has order $p^{r-\ell}$. By induction hypothesis, there is a normal subgroup $\overline{N}$ of $G/Z(G)$ of order $p^{k-\ell}$. Besides, by the correspondence theorem (4-th isomorphism theorem), we have $\overline{N}=N/Z(G)$ for some $N\unlhd G$. As we can see, $$|N|=|N/Z(G)||Z(G)|=p^{k-\ell}p^\ell=p^k.$$

Bernard Pan
  • 3,044