If $G$ is a group whose order is $p^n$($p$ is prime), then $G$ is solvable.

How am I going to show this? Any help is appreciated. Thank you.

  • 3
    $\begingroup$ Hint: The center of such a group is non-trivial by the class formula. $\endgroup$ Mar 18, 2013 at 17:52
  • 1
    $\begingroup$ See Rotman's or Rose's books, if you have an accesses to them. $\endgroup$
    – Mikasa
    Mar 18, 2013 at 17:54
  • $\begingroup$ Ironically the trigger that this post might get closed is exactly due to a good answer being added. $\endgroup$ Apr 5, 2019 at 2:14
  • $\begingroup$ $p$-groups are nilpotent, hence solvable (see last answer). $\endgroup$ Feb 12, 2020 at 12:10

2 Answers 2


Try by induction on the power of $p$. If $n=1$, $G$ is solvable by definition as a cyclic group of prime order.

Suppose that statement is true for all $k\leq n-1$. Suppose $|G|=p^n$. By the class equation, the center $Z(G)$ is nontrivial. So $Z(G)$ is normal in $G$ and abelian, hence solvable.

So either $G/Z(G)$ is a $p$-group of smaller order, or it is trivial.

The key theorem to remember is that if $H\unlhd G$ and $H$ is solvable and $G/H$ is solvable, then $G$ is also solvable. If $|G/Z(G)|< p^n$, then by induction $G/Z(G)$ is solvable, so $G$ is solvable. Otherwise you just have $G=Z(G)$.


An organized way to look at the problem is to have the following very important results in mind:

Proposition. Any nilpotent group is solvable.

Proof outline: We first refer to the following relation, valid for the descending central series of any arbitrary group $G$:

$$[\mathrm{C}^m(G), \mathrm{C}^n(G)] \subseteq \mathrm{C}^{m+n+1}(G)$$

for arbitrary $m, n \in \mathbb{N}$. On the basis of this, one can further establish by induction on $n \in \mathbb{N}$ that:

$$\mathrm{D}^n(G) \subseteq \mathrm{C}^{2^n-1}(G)$$

thanks to which matters become clear (if the $r$-th lower central subgroup is trivial, then $\mathrm{D}^r(G) \leqslant \mathrm{C}^{2^r-1}(G) \leqslant \mathrm{C}^r(G)=\{1_G\}$, bearing in mind the inequality $2^r \geqslant r+1$ valid for all $r \in \mathbb{N}$). $\Box$

Theorem. Let $p \in \mathbb{N}^*$ be a prime. Any $p$-group is nilpotent.

Proof outline: For arbitrary group $G$ convene to denote by $$\mathscr{S}(G)=\{H \subseteq G\ |\ H \leqslant G\}$$

the subset of all subgroups of $G$. If $m, n$ are natural numbers, then by $[m, n]$ we shall mean the interval between the two given by the order on $\mathbb{N}$ (so for instance $[3, 5]=\{3, 4, 5\}$).

One establishes by induction on $n \in \mathbb{N}$ that given group $G$ such that $|G|=p^n$ then there exists a finite sequence $H \in \mathscr{S}^{[0, n]}$ such that:

  1. $H_0=G$ and for any $k<n$ we have $H_k \geqslant H_{k+1}$ together with $|H_k:H_{k+1}|=p$ (in other words, the sequence is strictly decreasing).
  2. For any $k<n$ we have $[G, H_k] \leqslant H_{k+1}$.

To be noted that these conditions automatically entail $H_{k} \trianglelefteq G$ (what one calls a normal series) and $|G:H_k|=p^k$ for any $k \leqslant n$, thus in particular $H_n=\{1_G\}$. In other words, such a sequence will clearly be a lower central series that reaches the trivial subgroup.

The inductive step is carried out by relying on the fact that if the exponent $n \geqslant 1$ (i.e. the group $G$ is not trivial) then it will have nontrivial center; as $\mathrm{Z}(G)$ is also a nontrivial $p$-group, we have $p|\ |\mathrm{Z}(G)|$ and hence there exists $a \in \mathrm{Z}(G)$ of order $p$. Then the group $H=\langle a \rangle$ is a normal (even central) subgroup by which you can factor out in order to apply the inductive hypothesis. $\Box$


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