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.
Best Answer
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:
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.