# If $G$ is a groupe such that $|G|=p^m k$, does $G$ has a subgroup of order $p^n$ with $n<m$.

If $G$ is a groupe such that $|G|=p^m k$, does $G$ has a subgroup $H$ s.t. $|H|=p^n$ with $n<m$ ?

I know that $G$ has a $p-$sylow subgroup, i.e. a group of order $p^m$.

I also know that $G$ has an element of order $p$ and thus a subgroup of order $p$ (in fact $\left<g\right>$ where $g^p=1$).

1) But for $1<n<m$, is there a group of order $p^n$ ?

2) By the way, does all $p-$group (i.e. a group of order $p^n$) are abelian ? (in a solution of an exercise, they use such a property but I’ve never seen such a result).

#### Solutions Collecting From Web of "If $G$ is a groupe such that $|G|=p^m k$, does $G$ has a subgroup of order $p^n$ with $n<m$."

Yes there is such a subgroup. For a proof look at Theorem 1 in https://en.wikipedia.org/wiki/Sylow_theorems#Proof_of_the_Sylow_theorems.

To answer your second question: not all $p$-groups are abelian. The simplest counterexample is $Q_8$, the quaternion group. It is, however, the case that all $p$-groups have nontrivial center. This is a nifty consequence of the class equation. For $P$ a group of prime power order $p^m$,
$$|P| = |Z(P)| + \sum_{i=1}^r[P:C_P(g_i)]$$
where $g_1,\dots,g_r$ are representatives of the non-central conjugacy classes of $P$. Since then $C_P(g_i) \neq P$ by definition, we must have $p$ dividing each term of the sum. And since $p$ divides $|P|$ it must also divide $|Z(P)|$ so the center is nontrivial.

Let $P$ be the $p$-sylow sugroup of $G$ then $|P|=p^{m}$, there existe a normal subgroup $N$ of $P$ of order $p^{m-1}$, also there existe a normal subgroup of N of order $p^{m-2}$, ….

All those groups are subgroups of $G$