# Solvability of a group with order $p^n$

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.

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

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 $$k we have $$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 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. $$\Box$$