All $p$-groups have a normal subgroup of each possible order. [duplicate]

Solution 1:

In fact we can show that $G$ has a normal series with quotients $\mathbb{Z}/p$. By induction on $n$.

The step $n\to n+1$ is the only nontrivial one. Let $G$ be a group of order $p^{n+1}$.

It's enough to find a normal subgroup of $G$ of order $p$. It is a fact ( see the proof in the appendix below) that $Z(G)$ --the center of $G$, is nontrivial $|Z(G)| = p^k$ with $1\le k \le n+1$. Take any nontrivial element $z$ of $Z(G)$, $\text{ord}z = p^l$. Then $z' \colon = z^{p^{l-1}}$ has order $p$. The subgroup $H$ generated by $z'$ has order $p$ and being inside $Z(G)$ is necessarily normal. Consider the group $G/H$ of order $p^n$. By induction, there exists an increasing sequence $(\bar G_i)_{0\le i \le n}$ of normal subgroups of $G/H$ of orders $|\bar G_i| = p^i$. Their preimages $G_i$ in $G$ form an increasing sequence of normal subgroups of $G$ of orders $|G_i| = p^{i+1}$. Add to this $G_{-1} = (e)$

Proof that the center is nontrivial: Consider a finite $p$ group acting on a finite set $X$. Then cardinality of the set of fixed points of the action satisfies

$$|X_0| \equiv |X|\ ( \! \! \mod p\,)$$

This because the nontrivial orbits have cardinality $\equiv 0\ ( \! \! \mod p\,)$

Now consider the action of $G$ a finite $p$-group on $X=G$ by conjugation. The set of fixed points is $Z(G)$ and $|Z(G)|\equiv |G| (\mod p)$. But $|Z(G)| \ge 1$ since $e \in Z(G)$.