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.

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.

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.

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 $$i^2=j^2=k^2=ijk=-1.$$