# P-groups

Let $p$ be a positive prime number. A **p-group** is a group in which every element has order equal to a power of $p.$ A finite group is a $p$-group if and only if its order is a power of $p.$

There are many common situations in which $p$-groups are important. In particular, the Sylow subgroups of any finite group are $p$-groups. Since $p$-groups have many special properties, they are easier to understand and classify than arbitrary groups, but they are useful since they are, in a sense, the "building blocks" for arbitrary groups via the Sylow theorems.

## First properties

The first Sylow theorem can be used to show the following claim from the introduction:

A finite group is a $p$-group if and only if its order is a power of $p.$

First, suppose that the order of a finite group is a power of $p.$ By Lagrange's theorem, the order of any element divides the order of the group; but the only divisors of a power of $p$ are powers of $p,$ so the group is a $p$-group.

On the other hand, suppose that a group is a $p$-group. Suppose its order is not a power of $p;$ then there is another prime $q$ which divides its order. By the first Sylow theorem, there is a nontrivial Sylow $q$-subgroup, whose order is a power of $q.$ But then the order of any nontrivial element is a power of $q,$ which is not a power of $p,$ which is a contradiction. This proves the other direction of the equivalence.

## Nontrivial center and solvability

An elementary application of the class equation gives the following nontrivial fact about finite $p$-groups:

The center of a nontrivial finite $p$-group is nontrivial.

Recall that the **center** of a group $G$ is the set of elements $x$ which commute with everything in $G:$ $xg = gx$ for all $g\in G.$

This follows from the second form of the class equation: $|G| = |Z(G)| + \sum_{i=1}^\ell [G:C_G(g_\ell)]$ for some elements $g_1,\ldots,g_\ell$ not in the center. The point is that the terms $[G:C_G(g_\ell)]$ in the sum cannot equal $1,$ because if they did, $C_G(g_\ell)$ would be all of $G,$ and so $g_\ell$ would be in the center. If $G$ is a $p$-group, then those terms have to be divisible by $p$ (powers of $p,$ actually), and so taking both sides of the equation modulo $p$ gives $0 \equiv |Z(G)|.$ So $p$ divides $|Z(G)|,$ and hence it is not equal to 1.

Every $p$-group is solvable.

First there is a basic fact:

If $N$ and $G/N$ are solvable, so is $G.$

To see this fact, note that a composition series $1 \triangleleft H_1 \triangleleft H_2 \triangleleft \cdots \triangleleft G/N$ corresponds to a series $N \triangleleft K_1 \triangleleft K_2 \triangleleft \cdots \triangleleft G$ via the third isomorphism theorem; and the composition factors are the same. Concatenating with the series $1 \triangleleft N_1 \triangleleft N_2 \triangleleft \cdots \triangleleft N$ that exhibits the solvability of $N$ gives a series that exhibits the solvability of $G.$

Now the theorem follows by an easy induction. Suppose it is true for all $p$-groups of order < $|G|.$ (The base case, the trivial group, is trivially solvable.) If $G$ is abelian, then it is clearly solvable (since the composition factors are simple and abelian, hence cyclic of prime order). Otherwise, both $Z(G)$ and $G/Z(G)$ are smaller $p$-groups than $G$ (since $Z(G)$ is nontrivial by the previous theorem), and hence must be solvable by the inductive hypothesis. Then $G$ is solvable by the fact proved in the previous paragraph.

The nontriviality of the center is the basis for many inductive arguments. The following problem gives another example.

Let $G$ be a group of order $p^5,$ where $p$ is prime. Consider the following statements:

I. $G$ has a normal subgroup of order $p.$

II. $G$ has a normal subgroup of order $p^2.$

III. $G$ has a normal subgroup of order $p^3.$

IV. $G$ has a normal subgroup of order $p^4.$

How many of the statements I-IV is/are **always** true for any $G$ of order $p^5$?

**Terminology**: The *order* of a finite group is the number of elements in the group.

## Classification up to order $p^3$

Every group of prime order $p$ is isomorphic to the cyclic group ${\mathbb Z}_p.$

Let $g$ be a nontrivial element of $G.$ The order of $g$ is a nontrivial divisor of $p,$ so it must be $p.$ It is not hard to check that the function ${\mathbb Z}_p \to G$ defined by $n \mapsto g^n$ is an isomorphism.

Every group of order $p^2,$ where $p$ is a prime, is abelian. There are two such groups: ${\mathbb Z}_{p^2}$ and ${\mathbb Z}_p \times {\mathbb Z}_p.$

Let $G$ be a group of order $p^2.$ Every subgroup has order either $1,p,$ or $p^2$ by Lagrange's theorem. Suppose $G$ is not abelian. Since $Z(G)$ is nontrivial and not all of $G,$ it must have order $p.$ Let $y$ be an element not in $Z(G).$ Then $C_G(y)$ contains the center, but also contains $y.$ So it is strictly larger than a subgroup of order $p.$ The only possibility for its order is $p^2,$ but then everything commutes with $y$ and $y$ is in the center; contradiction.

The classification of abelian groups of order $p^2$ follows from the general classification theorem for abelian groups.

Groups of order $p^3$ are harder to classify. Here is the statement of the result, without a complete proof.

For any prime $p,$ there are five groups of order $p^3.$ Three of these are abelian: ${\mathbb Z}_{p^3}, {\mathbb Z}_{p^2} \times {\mathbb Z}_p, {\mathbb Z}_p \times {\mathbb Z}_p \times {\mathbb Z}_p.$ When $p=2,$ the two nonabelian groups are the dihedral group $D_4$ of rotations of the square, and the quaternion group $Q_8.$ When $p$ is odd, the two nonabelian groups are $H_p = \left\{ \begin{pmatrix} 1&a&b\\0&1&c\\0&0&1 \end{pmatrix} : a,b,c \in {\mathbb Z}_p \right\}$ and $A_p = \left\{ \begin{pmatrix} a&b \\ 0&1 \end{pmatrix} : a,b \in {\mathbb Z}_{p^2}, a \equiv 1 \pmod p \right\}.$ (In both cases, the group operation is matrix multiplication.)

**Notes:**

(1) When $p=2,$ $H_2 \cong A_2 \cong D_4.$ For odd $p,$ they are not isomorphic: every nontrivial element of $H_p$ has order $p,$ but $A_p$ has elements like $\begin{pmatrix} 1&1\\0&1 \end{pmatrix}$ of order $p^2.$

(2) $H_p$ and ${\mathbb Z}_p \times {\mathbb Z}_p \times {\mathbb Z}_p$ both have one element of order $1$ and $p^3-1$ elements of order $p,$ but they are not isomorphic. So just knowing the orders of all the elements in a group is not enough to determine the group up to isomorphism.

(3) $D_4,$ the dihedral group of a square, is the group of rotations and reflections of the square. $Q_8$ is the group whose elements are $\pm 1, \pm i, \pm j, \pm k$ with multiplication given by $i^2=j^2=k^2=ijk=-1.$