Classification of finite p-groups with cyclic maximal subgroup

This page classifies all finite $$p$$-groups that possess a cyclic maximal subgroup.

For odd primes
Let $$p$$ be an odd prime, and $$G$$ be a group of order $$p^n$$ with a cyclic maximal subgroup $$M$$ (so $$M \cong \mathbb{Z}/p^{n-1}\mathbb{Z}$$). Then, there are two possibilities for $$G$$:


 * 1) $$G$$ is isomorphic to $$\mathbb{Z}/p^{n-1}\mathbb{Z} \times \Z/p\mathbb{Z}$$, with $$M$$ being the first direct factor (we could choose other isomorphisms where $$M$$ is not one of the direct factors).
 * 2) $$G$$ is isomorphic to $$\mathbb{Z}/p^n\mathbb{Z}$$, with the isomorphism sending $$M$$ to the subgroup of multiples of $$p$$
 * 3) $$G$$ is an internal semidirect product of the subgroup $$M$$ and a cyclic group of order $$p$$, whose generator acts on $$M$$ via multiplication by $$p^{n-2} + 1$$.

When $$n = 2$$, then (2) and (3) are equivalent, for $$n \ge 3$$, we get three possible isomorphism classes for $$G$$ from the above.

For the prime 2
Let $$G$$ be a group of order $$2^n$$ with a cyclic maximal subgroup $$M \cong \mathbb{Z}/2^{n-1}\mathbb{Z}$$. Then, there are six possibilities for $$G$$:


 * 1) $$G$$ is isomorphic to $$\mathbb{Z}/2^{n-1}\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$, with $$M$$ being the first direct factor (we could choose other isomorphisms where $$M$$ is not one of the direct factors).
 * 2) $$G$$ is isomorphic to $$\mathbb{Z}/2^n\mathbb{Z}$$, with the isomorphism sending $$M$$ to the subgroup of multiples of $$p$$
 * 3) $$G$$ is a dihedral group: it is a semidirect product of $$M$$ and a cyclic group of order two, which acts on $$M$$ via multiplication by -1.
 * 4) $$G$$ is a semidirect product of $$M$$ and a cyclic group of order two, which acts on $$M$$ via multiplication by $$2^{n-2} + 1$$.
 * 5) $$G$$ is a semidirect product of $$M$$ and a cyclic group of order two, which acts on $$M$$ via multiplication by $$2^{n-2} - 1$$
 * 6) $$G$$ is a generalized quaternion group

Generalizations

 * Classification of metacyclic p-groups
 * Classification of finite p-groups with self-centralizing cyclic normal subgroup

Proof for odd primes
We first outline a naive proof, after which we see a slicker proof using the ideas of second cohomology of cyclic groups.

Since $$G$$ is a $$p$$-group, it is nilpotent, so any subgroup of index $$p$$ is normal. Thus, $$M \triangleleft G$$. Pick a $$a \in G \setminus M$$. Then, conjugation by $$a$$ induces an automorphism $$\sigma$$ of $$M$$. Moreover, $$a^p \in M$$, because $$M$$ has index $$p$$, so $$\sigma^p$$ is conjugation by some element of $$M$$. Since $$M$$ is cyclic and hence Abelian, $$\sigma^p$$ must therefore be the identity map. Thus $$\sigma$$ is an automorphism of $$M$$ that has order either 1 or $$p$$.

Case that $$\sigma$$ has order one
In this case, $$G$$ is generated by $$M$$ and an element $$a$$ that commutes with $$M$$, so $$G$$ is Abelian. (For convenience we use multiplicative notation in $$M$$). There are two possibilities:


 * 1) $$a^p$$ is not a generator for $$M$$: Then, by the structure of cyclic groups there exists $$b \in M$$ such that $$b^p = a^p$$. Then the subgroup generated by $$(ab^{-1})$$ is cyclic of order $$p$$, and $$G$$ is an internal direct product of $$M$$ with this cyclic subgroup.
 * 2) $$a^p$$ is a generator for $$M$$: In this case, $$G$$ is cyclic of order $$p^n$$

Case that $$\sigma$$ has order $$p$$
In this case, the subgroup generated by $$\sigma$$ is a subgroup of order $$p$$ in the multiplicative group mod $$p^{n-1}$$. But there's only one such subgroup: the multiplicative subgroup comprising multiplications by $$kp^{n-2} + 1$$. Replacing $$a$$ by a suitable power thereof, we can assume that $$\sigma$$ acts as multiplication by $$p^{n-2} + 1$$.

Further, it is clear that $$a^p$$ cannot be equal to a generator of $$M$$, because otherwise $$a$$ would commute with $$M$$ making the group Abelian and $$\sigma$$ trivial. Thus, there exists $$b \in M$$ such that $$a^p = b^p$$.

Now consider the element $$ab^{-1}$$. We want to show that $$(ab^{-1})^p$$ is the identity. First, let's compute the commutator of $$a$$ and $$b^{-1}$$:

$$[a,b^{-1}] = ab^{-1}a^{-1}b = b^{-p^{n-2}} = a^{-p^{n-2}}$$

Thus, the commutator has order $$p$$ and commutes with both $$a$$ and $$b$$.

Now, use the fact that since the commutator of $$a$$ and $$b$$ commutes with both $$a$$ and $$b$$, the subgroup they generate is nilpotent of class two, so using the formula for powers of product in group of class two, we have:

$$a^p b^{-p} = [a,b^{-1}]^{p(p-1)/2} (ab^{-1})^p$$

Since the commutator has order $$p$$, its $$p(p-1)/2$$th power is the identity, so the power of the commutator is trivial. Further, since $$a^p = b^p$$, the left side is trivial, so we get $$(ab^{-1})^p$$ is trivial.

Thus, the element $$ab^{-1}$$ generates a cyclic subgroup of order $$p$$, whose internal semidirect product with $$M$$ is $$G$$.

Proof at the prime 2
The main difference between the prime 2 and the odd primes is the fact that the multiplicative group mod $$2^{n-1}$$ is not cyclic, and thus it has four possible squareroots of 1. Hence, there are four different possible actions that the quotient $$G/M$$ can have on $$M$$, and we need to classify all possible groups for each of the actions.

The other main difference is that while for odd primes $$p$$, $$p$$ divides $$p(p-1)/2$$, this isn't true for 2, so applying the formula for product of powers yields weaker results.

As in the previous situation, let $$a$$ be an element in $$G \setminus M$$, let $$\sigma$$ be the automorphism of $$M$$ induced via conjugation by $$a$$. As before, $$a^2 \in M$$, and $$M$$ is Abelian, so $$\sigma^2$$ is the identity on $$M$$.

Case that $$\sigma$$ has order 1
In this case, $$a$$ commutes with $$M$$, so $$G$$ is Abelian. There are two possibilities:


 * 1) $$a^2$$ is not a generator for $$M$$. Then $$a^2 = b^2$$ for some $$b \in M$$, so $$(ab^{-1})^2$$ is the identity. Then $$G$$ is an internal direct product of $$M$$ and the subgroup generated by $$ab^{-1}$$
 * 2) $$a^2$$ is a generator for $$M$$. Then, $$G$$ is cyclic of order $$2^n$$.

Case that $$\sigma$$ is the inverse map
If $$\sigma$$ has order 2, it must be an element of order 2 in the automorphism group of $$\mathbb{Z}/2^{n-1}\mathbb{Z}$$. There are three such elements: multiplication by $$-1$$, multiplication by $$2^{n-1} + 1$$ and multiplication by $$2^{n-2} - 1$$. (since we'll use multiplicative notation for $$M$$, we'll effectively be talking of raising to the powers of the numbers).

Let's first consider the case that $$\sigma$$ acts as multiplication by $$-1$$. In multiplicative notation, $$\sigma(x) = x^{-1} \ \forall \ x \in M$$. Our goal is now to answer the question: can we find some element in the coset of $$a$$ that has order 2 in $$G$$? If not, can we still find some element of small order? We try to repeat the procedure we followed last time.

Clearly $$a^2 \in M$$ but $$a^2$$ is not a generator of $$M$$, because otherwise $$G$$ would be cyclic. So there exists $$b \in M$$ such that $$a^2 = b^2$$.

Using group cohomology for odd primes
Let $$C$$ denote the cyclic group of order $$p$$ and $$M$$ denote the cyclic group of order $$p^k$$, where $$k = n -1$$. We want to classify all groups of order $$p^n$$ with a normal subgroup isomorphic to $$M$$ and quotient group isomorphic to $$C$$.

Classifying the possible actions
The first step behind classification is to find all elements of:

$$\operatorname{Hom}(C, \operatorname{Aut}(M))$$

Let's do this. By definition, $$\operatorname{Aut}(M)$$ is a cyclic group of order $$p^{k-1}(p-1)$$ and the Sylow $$p$$-subgroup is obtained as multiplications by $$mp + 1$$ for integers $$m$$. The only elements of order $$p$$ in this are elements of the form $$mp^{k-1} + 1$$. Thus, there are $$p$$ possible homomorphisms from $$C$$ to $$\operatorname{Aut}(M)$$, the homomorphisms sending the generator to $$mp^{k-1} + 1$$ for different possible values of $$m \in \{ 0,1,2,\dots,p-1\}$$.

Further, the homomorphisms fall into two equivalence classes modulo automorphisms of $$C$$: the trivial homomorphism, and the homomorphism sending the generator to multiplication by $$p^{k-1} + 1$$. Thus, it suffices to classify all congruence classes of extensions for these two actions.

Computing congruence classes of extensions for the trivial action
We first compute the second cohomology group for the trivial homomorphism.

Let us apply the result discussed in the second cohomology of cyclic groups. Here, the action is trivial, so the feasible cocycles are given by:

$$\{ a \in M \mid a = a \}$$

which is clearly the whole of $$M$$. On the other hand, the coboundaries are given by (the sum here is $$p$$ times):

$$\{ a + a + \dots + a \mid a \in M \}$$

which is the subgroup $$pM$$ of multiples of $$p$$.

Thus, the second cohomology group is $$M/pM = \mathbb{Z}/p\mathbb{Z}$$. A concrete interpretation of this:


 * The zero element in this cohomology group corresponds to the direct product of $$M$$ and $$C$$.
 * Any nonzero element in this cohomology group corresponds to the cyclic group of order $$p^{k+1}$$. All these are related by automorphisms of the extension.

Thus, up to automorphisms, there are only two congruence classes of extensions for the trivial action: the direct product, and the cyclic group.

Computing congruence classes of extensions for the nontrivial action
Using the results discussed in second cohomology of cyclic groups, it follows that the second cohomology group is given as the following quotient:

$$\frac{\{ a \in M \mid (p^{k-1} - 1)a = a \}}{\{ a + (p^{k-1} + 1)a + (2p^{k-1} + 1)a + \dots + (p^{k-1}(p-1) + 1)a \mid a \in A \}}$$

The expression in the numerator is the group of feasible cocycles, and the expression in the denominator is the group of coboundaries. Clearly, the expression in the numerator is $$pM$$, and the expression in the denominator is also $$pM$$, so the quotient is trivial. In other words, the second cohomology group is trivial, and thus, every extension corresponding to this action splits.

Using group cohomology for the prime 2
Let $$C$$ denote the cyclic group of order $$2$$ and $$M$$ denote the cyclic group of order $$2^k$$, where $$k = n-1$$. We want to classify all groups of order $$2^n$$ with a normal subgroup isomorphic to $$M$$ having quotient group isomorphic to $$C$$.

Classifying the possible actions
The actions of $$C$$ on $$M$$ are given by elements of the set:

$$\operatorname{Hom}(C,\operatorname{Aut}(M))$$

Now, $$\operatorname{Aut}(M)$$ is a direct product of a cyclic group of order $$2^{k-2}$$ and a cyclic group of order $$2$$. It has three elements of order two, as well as the identity element, so there are four possible homomorphisms from $$C$$ to this group. The four elements are:


 * The identity map.
 * The multiplication by $$-1$$ map (or the inverse map).
 * The multiplication by $$2^{k-1} - 1$$ map.
 * The multiplication by $$2^{k-1} + 1$$ map.

The second cohomology group for the trivial action
We first compute the second cohomology group for the trivial homomorphism.

Let us apply the result discussed in the second cohomology of cyclic groups. Here, the action is trivial, so the feasible cocycles are given by:

$$\{ a \in M \mid a = a \}$$

which is clearly the whole of $$M$$. On the other hand, the coboundaries are given by:

$$\{ a + a \mid a \in M \}$$

which is the subgroup $$2M$$. Thus, the quotient, which is the second cohomology group, is cyclic of order two:


 * The trivial element corresponds to the direct product of $$M$$ and $$C$$.
 * The nontrivial element corresponds to a cyclic group of order $$p^{k+1}$$.

The second cohomology group for the inverse map
The second cohomology group is given by:

$$\frac{\{ a \in M \mid -a = a \}}{ \{ a - a \mid a \in M \}}$$

The group on top is a subgroup of order two, and the group at the bottom is trivial, so the quotient is cyclic of order two. Thus, there are two such groups. These are described specifically as the dihedral group of order $$2^n$$ and the generalized quaternion group of order $$2^n$$.

The second cohomology groups for $$2^{k-1} \pm 1$$
There are both trivial. The second cohomology group for multiplication by $$2^{k-1} +1$$ is given by:

$$\frac{\{ a \in M \mid (2^{k-1} + 1)a = a \}}{\{a + (2^{k-1} + 1)a = 0 \mid a \in M\}}$$

both the groups here are $$2M$$, so the quotient is trivial.

For multiplication by $$2^{k-1} - 1$$, the second cohomology group is given by:

$$\frac{\{ a \in M \mid (2^{k-1} - 1)a = a \}}{\{ a + (2^{k-1} - 1)a \mid a \in M \}}$$

Both the groups here are $$(2^{k-1})M$$, and the quotient is trivial.