Let $P$ a group of order $p^s$ ($p$ is prime) and $t \leq s$. Use Cauchy's Theorem and the proposition $7.2$ for showing P contains a subgroup of order $p^t$.
Cauchy theorem :
(1) Let $G$ a finite group et let $p$ a prime divisor of $\mid G \mid$. Then the group contain at least an element of order $p$.
(2) Let $n_p$ the number of elements in $G$ of order $p$ and let $N_p$ the number of subgroups of $G$ of order $p$. Then $n_p + 1$ and $N_p - 1$ are divisibles by $p$ and $n_p = N_p(p-1)$.
Proposition $7.2$ :
Let $P$ a finite group of order $p^r$, where $p$ is a prime number and $r>0$. The center $Z(P)$ of $P$ has at least $p$ elements. The order of $Z(P)$ isn't $p^{r-1}$. If $P$ isn't abelian, then $P/Z(P)$ is not cyclic.
I have seen that the stabilizer $P_x$ is a subgroup of $P$, and the orbit-stabilizer formula could be useful here.
I'm on this problem for a while now. Can anyone give me a little hint?
One can use the class equation for $G$ a finite group to prove the more general claim that if $p^s\mid |G|$, then $G$ contains a subgroup of order $p^s$. The class equation for $G$ reads
$$|G|=|Z(G)|+\sum |G:C(x_i)|$$
for finitely many $x_1,\ldots,x_r$ representatives of the nontrivial conjugacy classes of $G$. This is, I suppose, how you've proven that $Z(G)$ is nontrivial. We want to show that $G$ contains a subgroup of order $p^t$ for $t<s$.
If $G$ is abelian, by Cauchy we can take an element $g$ of order $p$ (we don't need $G$ abelian here). Then $G/(g)$ is a group of order $p^{s-1}$ and by induction it contains a subgroup of order $p^{t-1}$. This lifts to a subgroup of order $|(g)|p^{t-1}=p^t$ of $G$ by the correspondence of subgroups of $G$ and this quotient.
We can thus assume that $G$ is nonabelian, so that $Z(G)$ is proper in $G$, but still nontrivial. If $p^t$ divides $|Z(G)|$ then, since $Z(G)$ is abelian, we can use the above. So we can assume that $p^t$ doesn't divide $|Z(G)|$. It must then not divide $|G:C(x_i)|$ either, for some $i$. In this case, $p^t$ divides $|C(x_i)|$, and since $C(x_i)$ is a proper $p$-subgroup of $G$, we can apply induction on its order and obtain the desired subgroup of order $p^s$.