Prove that if $|G|=p^n$, for some prime number $p$, then there exists a normal subgroup of $G$ of order $p^m, m\le n $.

Question

Prove that if $|G|=p^n$, for some prime number $p$, then there exists a normal subgroup of $G$ of order $p^m, m\le n $.

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My first attempt was to use the proposition below, but I figure that the stabilizer is not a normal subgroup, and I wasn't sure if I can use this without having a set to act on.

Can I use the proposition that says if $|G|=p^n$ and $G$ acts on a set $X$ then $|X|=|X^G| \pmod{p}$, where $X^G=\{x\in X \mid \forall g\in G, gx=x\}$

On $G$ itself ?

Meaning, can I write $ \def\st{\operatorname{st}} |G|=|\st(x)|\pmod{p}$? Where $\st(x)$ I denote the stabilizer of $G$

My second attempt was to use the fact that the center is a normal subgroup and maybe that will do it by using: $$ |G|=|Z(G)| \;{}+ \sum_{|o(x)|>1} \frac{|G|}{|\st(x)|} $$

Where $o(x)$ is the orbit and $Z(G)$ is the center.

Answer 1

Work by induction on $m$. Assume $n \gt 1$. First note that the class equation assures you that $Z(G)$ is non-trivial, so it must have an element of order $p$, which in turn generates a normal (because central) subgroup $N$ of order $p$. In other words, any such group must have a normal subgroup of order $p^1=p$. That's the base case for our induction.

Now consider $G/N$. If $1 \lt m \lt n$, then $\vert G/N \vert = p^{n-1} \gt p^{m-1} \gt 1$ and by our inductive hypothesis, $G/N$ has a normal subgroup of order $p^{m-1}$, which corresponds to a normal subgroup of $G$ containing $N$ with order $p^m$.

Answer 2

$ \def\st{\operatorname{st}} |G|=|\st(x)|\pmod{p} \Rightarrow |\st(x)|=p^{i_x},i_x\in \{1,...,n-1\}$

$$ |G|=|Z(G)| \;{}+ \sum_{|o(x)|>1} \frac{|G|}{|\st(x)|} $$

$$ p^n=|Z(G)|+\sum_{|o(x)|>1}p^n/p^{i_x}$$

$$ |Z(G)|=p^m\left(p^{n-m}-\sum_{|o(x)|>1}p^{n-m}/p^{i_x}\right),$$ where $m=\min\{i_x\}$.

Thus $$p^m\mid |Z(G)|,$$

where $o(x)$ is the orbit and $Z(G)$ is the center.