The structure theory of finite p-groups does not stop with the nontrivial center. The focus keyword is subgroups of finite p-groups, because finite p-groups contain subgroups of every possible p-power order below their own order. This is a strong internal existence theorem and a conceptual bridge from Cauchy's theorem to Sylow theory. Before finding large p-subgroups inside arbitrary finite groups, we first learn how p-subgroups grow inside p-groups. :::definition[Subgroup Chain in a p-Group] Let $G$ be a finite p-group of order $p^n$. A subgroup chain in a p-group is a sequence $$\begin{equation*} \{e\}=H_0\leq H_1\leq\cdots\leq H_n=G \end{equation*}$$ such that $|H_i|=p^i$ for every $i$ satisfying $0\leq i\leq n$. ::: :::lemma Let $G$ be a nontrivial finite p-group. Then $G$ has a normal subgroup of order $p$. ::: :::proof Given that $G$ is a nontrivial finite p-group. To prove that $G$ has a normal subgroup of order $p$. By the nontrivial center theorem, $Z(G)$ is nontrivial. Since $Z(G)$ is a finite p-group, Cauchy's theorem gives an element $z\in Z(G)$ such that $o(z)=p$. Let $N=\langle z\rangle$. Then $|N|=p$. Since $z\in Z(G)$, every element of $N$ commutes with every element of $G$. Therefore $N\leq Z(G)$. Every subgroup of the center is normal in $G$. Hence $G$ has a normal subgroup of order $p$. $\square$ ::: :::theorem Let $G$ be a finite p-group of order $p^n$. Then for every integer $i$ satisfying $0\leq i\leq n$, the group $G$ has a subgroup of order $p^i$. ::: :::proof Given that $G$ is a finite p-group of order $p^n$. To prove that for every integer $i$ with $0\leq i\leq n$, the group $G$ has a subgroup of order $p^i$. We use induction on $n$. If $n=0$, then $G=\{e\}$, and the only required subgroup has order $1=p^0$. Assume the result holds for all finite p-groups of order smaller than $p^n$, where $n\geq1$. By the lemma, $G$ has a normal subgroup $N$ of order $p$. Then $G/N$ is a finite p-group of order $p^{n-1}$. Let $i$ satisfy $0\leq i\leq n$. If $i=0$, then $\{e\}$ has order $1=p^0$. If $i\geq1$, then $i-1$ satisfies $0\leq i-1\leq n-1$. By the induction hypothesis applied to $G/N$, there is a subgroup $K/N\leq G/N$ such that $|K/N|=p^{i-1}$. The inverse image $K$ is a subgroup of $G$ containing $N$, and $$\begin{align*} |K|&=|K/N||N|\\ &=p^{i-1}\cdot p\\ &=p^i. \end{align*}$$ Hence $G$ has a subgroup of order $p^i$ for every $0\leq i\leq n$. $\square$ ::: :::theorem Let $G$ be a finite p-group and let $H\leq G$. If $[G:H]=p$, then $H$ is normal in $G$. ::: :::proof Given that $G$ is a finite p-group and $H\leq G$ with $[G:H]=p$. To prove that $H$ is normal in $G$. Let $G$ act by left multiplication on the set $G/H$ of left cosets of $H$. This action gives a homomorphism $\varphi:G\to S_p$. The kernel of this action is contained in $H$, because any element fixing the coset $H$ must lie in $H$. The image $\varphi(G)$ is a p-subgroup of $S_p$, so $|\varphi(G)|$ divides both $|G|$ and $p!$. The highest power of $p$ dividing $p!$ is $p$, so $|\varphi(G)|$ is $1$ or $p$. The action on $G/H$ is transitive, so $p$ divides $|\varphi(G)|$. Therefore $|\varphi(G)|=p$. Hence $[G:\ker\varphi]=p$. Since $\ker\varphi\leq H$ and $[G:H]=p$, it follows that $\ker\varphi=H$. Since kernels of homomorphisms are normal, $H$ is normal in $G$. Hence $H\trianglelefteq G$. $\square$ ::: :::example Let $G=\mathbb{Z}_{27}$ under addition modulo $27$. Since $|G|=27=3^3$, the theorem guarantees subgroups of orders $1,3,9,27$. These are $$\begin{equation*} \{0\},\quad \langle9\rangle,\quad \langle3\rangle,\quad \mathbb{Z}_{27}. \end{equation*}$$ Hence $\mathbb{Z}_{27}$ has a full subgroup chain. ::: This preview displays the subgroup orders guaranteed inside a finite p-group of order $p^n$. Choose $p$ and $n$, then follow the chain from the trivial subgroup to the whole group. Each step multiplies the order by $p$, matching the induction proof through a normal subgroup of order $p$. Try the presets for $\mathbb Z_{27}$ and $\mathbb Z_{16}$, then change $n$ to see how the chain grows. :::scientific-preview[Subgroup Chain Builder]