For a prime number $p$, a $p$-nilpotent finite group can be defined as a finite group $G$ with the two following properties :
a) the elements of $G$ with order not divisible by $p$ form a subgroup of $G$;
b) this subgroup is of order $\vert G \vert / p^{n}$, where $p^{n}$ denotes the greatest power of $p$ dividing $\vert G \vert .$
In fact, property b) is a consequence of property a), but I can only prove this by use of a theorem of Frobenius that I don't find in the textbooks of Scott, Rotman, Rose and Robinson. It is stated and proved in Zassenhaus' "The Theory of Groups" (repr. Dover), p. 28. The proof is elementary, but not very easy.
This theorem of Frobenius states that if $G$ is a finite group, if $d$ is a natural divisor of $\vert G \vert $, then the elements $x$ of $G$ such that $x^{d} = 1$ are in number divisible by $d$. (In fact, the theorem is a little stronger.)
This theorem can be used in the following manner in order to prove that a) implies b). By a), the elements of $G$ with order not divisible by $p$ form a subgroup of $G$. Let $H$ denote this subgroup. If we put $d = \vert G \vert / p^{n}$, where $p^{n}$ denotes the greatest power of $p$ dividing $\vert G \vert $, then $H$ is the set of all elements of $G$ such that $x^{d} = 1$, thus, by the theorem of Frobenius,
(1)$\qquad$ the order of $H$ is divisible by $d$, i.e. by $\vert G \vert / p^{n}$.
On the other hand, since every element of $H$ has order coprime to $p$, the order of $H$ is coprime to $p$; thus, by (1) (and since $\vert H \vert$ divides $\vert G \vert$), $\vert H \vert = \vert G \vert / p^{n}$, which proves b).
Since the proof of Frobenius' theorem is not very easy and this theorem is not a favorite of the authors of textbooks, my question is : is it is possible to prove that a) implies b) in a simple manner, without using Frobenius' theorem ? Thanks in advance for the answers.
Best Answer
Here is a proof that a) implies b). I think you just need Sylow's theorem. Let $p$ be a prime number.
Let $G$ be a finite group of order $np^{\alpha}$, where $p \nmid n$.
Suppose that the set $H = \{x \in G : p \nmid o(x) \}$ forms a subgroup in $G$. Then certainly $|H|$ divides $n$, because otherwise $|H|$ would be divisible by $p$, and thus would contain an element of order $p$ by Cauchy's theorem.
Now for any prime $q$ dividing $n$, let $Q$ be a Sylow $q$-subgroup in $G$. Then $Q \leq H$, so $|H|$ is divisible by $|Q|$. Thus conclude that $n$ divides $|H|$, so in fact $|H| = n = |G|/p^{\alpha}$.