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.
 A: Let $\rm P$ be a subgroup property (e.g. proper, nontrivial, normal, central, characteristic, finite-index, and so on). I picked up this cool term from GroupProps. Then:


*

*a "$\rm P$ maximal subgroup" is a maximal subgroup which is $\rm P$, whereas

*a "maximal $\rm P$ subgroup" is a subgroup which is maximal among $\rm P$ subgroups.


The set of all subgroups of a group $G$ forms a partially ordered set ${\cal L}(G)$ under inclusion (in fact it forms a lattice). Any subset $X\subseteq{\cal L}(G)$ inherits the partial order and becomes a poset in its own right, but $X$ may not contain any maximal elements. An element of a poset is maximal if it is not less than any other element. In general there may be no maximal elements or multiple maximal elements, and a maximal element may not be comparable with every other element.
If $G$ is finite though, then ${\cal L}(G)$ is finite and so is the set of $\rm P$ groups (whatever the property $\rm P$ is), and thus there must be at least one maximal element. So there is a subgroup of $G$ which is maximal among $\rm P$ groups. Of course this situation is ${\rm P}=$ being a normal subgroup.
A: 
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. $\square$
A: This is an understandable misreading of the statement. The meaning of

Let $K$ be a maximal normal subgroup of $G$ contained in $H$

is: Let $K$ be a normal subgroup of $G$ contained in $H$, which is maximal with respect to this property. Informally, "let $K$ be one of the largest among the normal subgroups of $G$ that are contained in $H$". More formally: If $\mathcal S$ denotes the collection of all subgroups of $H$ that are normal in $G$, let $K$ an element of $\mathcal S$ that is maximal with respect to inclusion.
