about a maximal normal subgroup of a $p$ group. i'm studying bhattacharya's basic algebra. it introduces the concept of the group action in chapter 4 and proves the class equation. and derives simple properties of $p$ group using the equation. the author proves the thm stating :
"Let $G$ be a finite group of order $p^{n}$, where $p$ is a prime $n$ is a natural number. If $H$ is a proper subgroup of $G$, then $H$ is a properly contained in $N(H)$; hence, if $H$ is a subgroup of order $p^{n-1}$, then $H$ is normal in $G$".
and the author starts the proof by saying: " Let $K$ be a maximal normal group of $G$ contained in $H$......"
Well how do you know $H$ contains a maximal normal group of $G$? This seems nontrivial to me......any help will be appreciated.


If $H$ is a proper subgroup of the finite group $G$ then there is a maximal subgroup of $G$ containing $H$.

Note that a subgroup M of $G$ is called maximal if $M \neq G$ and the only subgroups of $G$ that contain $M$ are $M$ and $G$) Let's prove the statement.
Proof. Consider the set $S=\{K\leq G: K\neq G, H\leq K\}$. This is a subset of $\mathcal{P}(G)$ and so is partially ordered by $\subseteq$ and is finite since $G$ is finite. If $S=\varnothing$, then $H$ itself is the maximal subgroup. Now suppose that $S$ is not empty and that no maximal subset containing $H$ exists. Then there is an element $K_{0} \in S$. Since $K_{0}$ is not maximal there exists an element $K_{1} \in S$, such that $K_{0} \leq K_{1}$ and $K_{0} \neq K_{1}$. Similarly since $K_{1}$ is not maximal there exists an element $K_{2} \in S$ such that $K_{0}\leq K_{1} \leq K_{2}$ and $K_{1} \neq K_{2}$. Thus, we may construct an infinite ascending chain of subgroups $K_{0}<K_{1}<K_{2}<\cdots$. This is a contradiction since $S$ is finite. Therefore, a maximal subgroup containing $H$ must exist. $\square$
