IV. SYLOW p-GROUPS AND p-GROUPS

§ 1. The Sylow Theorems

In a finite group of order N, the order of every subgroup is a divisor of N. On the other hand there need not be a subgroup with order d for every divisor d of N. For example, in the tetrahedral group, as one can see easily, there is no subgroup of order 6. We shall now prove, however, that for every power p^a of a prime dividing N there is a subgroup with the order p^a.

DEFINITION: A group is said to be a p-group if the order of each of its elements is a power of the prime p.

We determine the largest possible p-groups in the finite group .

DEFINITION: A subgroup of is said to be a Sylow p-group, if its order is equal to the greatest power of the natural prime p dividing N.

For example, the four group is a Sylow 2-group of the tetrahedral group. A Sylow p-group of is denoted by S_p or by . The normalizer of S_p in is denoted by N_p the center of S_p by Z_p.

THEOREM 1. For every natural prime p, every finite group contains a Sylow p-group.

Proof: If the order N of is 1, then the theorem is clear. Now let N > 1 and assume the theorem proven for groups of order smaller than N.

If in the center of there is an element a of order m.p, then the factor group / (a^m) is of order and contains by the induction assumption a Sylow p-group /(a_m) of order p^n-1, where is not divisible by p.

is of order pⁿ and therefore is a Sylow p-group of .

Now let there be no element of order divisible by p in the center of . If the order of were divisible by p, then the factor group of with respect to a cyclic normal subgroup (a) 1 is of order divisible by p. But then by the induction hypothesis /(a) would contain a Sylow p-group 1, and therefore would contain elements b(a) of order divisible by p. Then the order of b in would be divisible by p. Therefore the order of is not divisible by p. If p†N then e is the Sylow p-group sought. If p/N then it follows from the class equation

and from p † (3:1), p/N, that at least one h_i > 1 is not divisible by p. contains a normalizer N_i of index h_i > 1 and therefore N_i contains, by the induction hypothesis, a Sylow p-group . Since p † h_i, is also a Sylow p-group of .

COROLLARY: For every prime divisor p of the order of a finite group there is an element of order p (Cauchy).

The order and exponent of a finite group have the same prime divisors.

It is a p-group if and only if its order is a power of p.

THEOREM 2: If is a Sylow p-group of and a normal subgroup of , then is a Sylow p-group of ; / is a Sylow p-group of /.

Proof:¹ A subgroup of is a Sylow p-group if and only if

1. The order of is a power of p (written: pp),

2. The index of is prime to p (written: prime).

Now we may construct the diagram to the left and, observe first that : is prime to p, and : is a p-power. From the second isomorphism theorem it follows that : is a p-power, and : is prime to p, from which the theorem follows.

If a Sylow p-group is a normal subgroup of then it is the only Sylow p-group, since for every other Sylow p-group it ₁ follows that ₁ is of p-power order, but :₁ is prime to p; and therefore ₁ = = ₁. Consequently a Sylow p-group S_p of a finite group is the only Sylow p-group of its normalizer N_P.

THEOREM 3: All Sylow p-groups of a finite group are conjugate under . Their number when divided by p leaves a remainder 1.

Proof: Let the Sylow p-groups of be = ₁, …, _r.

Under the mapping of onto the group of inner automorphisms, is represented as a permutation group. Since conjugate subgroups have the same order, the _i are transformed into each other by , so that we obtain a representation Δ of as a permutation group of degree r. By a remark above, transforms only ₁ and no other _i into itself. Consequently there is only one system of transitivity of first degree. The other systems of transitivity of Δ have a degree > 1 which is a divisor of : 1 and which, therefore, is a p-power. Consequently r ≡ 1 (p).

transforms the s = :N_p Sylow p-groups conjugate to under among themselves, and as above it follows that s ≡ 1 (p). If there were another system of conjugate Sylow p-groups, then its members would be transformed into each other by in systems of transitivity whose degree would be divisible by p. The system would therefore contain a number s₁, divisible by p, of Sylow p-groups; on the other hand we conclude for s₁, just as we did for s, that s₁ ≡ 1 (p). Consequently all Sylow p-groups are conjugate to , Q.E.D.

THEOREM 4: Every p-group in is contained in a Sylow p-group.

Proof: We replace by in the proof of the previous theorem. Let the transformed objects again be ₁, …, _r. The degree of a system of transitivity of Δ is either 1 or a p-power. Since r ≡ 1 (p), there is certainly a system of transitivity of degree 1. Therefore there is a _i which is transformed into itself by all the elements of . Since _i is a p-group which contains _i, we have , Q.E.D.

THEOREM 5: Every subgroup of which contains the normalizer N_p of a Sylow p-group S_p, is its own normalizer.

Proof: We must show that x x^-1 implies

x .

In any case S_p and xS_px^-1 are Sylow p-groups of , and by Theorem 3 there is a U in such that U x S_p x^-1U^-1 = S_p;

therefore  U x N_p

therefore x   Q.E.D.

THEOREM 6: If the p-group contained in the finite group is not a Sylow p-group, then the normalizer N of is larger than .

Proof: If p † : N then the theorem is clear; if, however : N = pr, then transforms the pr subgroups conjugate to in systems of transitivity whose degrees are 1 or numbers divisible by p. Since is transformed into itself, there are at least p subgroups ₁ = , ₂, …, _p, conjugate to which are transformed into themselves by . Consequently N₂ is greater than ₂, and therefore N is greater than , Q.E.D.

COROLLARIES:

1. Every maximal subgroup of a p-group is a normal subgroup; therefore it is of index p.

2. If a p-group is simple then it is of order p.

3. The composition factors of a p-group are of order p and therefore every p-group is solvable.

§ 2. Theorems on Sylow p-Groups

Information on the intersection of different Sylow p-groups is given by

THEOREM 7: In the normalizer of a maximal¹ intersection of two different Sylow p-groups of we have:

1. Every Sylow p-,group of N contains properly.

2. The number of Sylow p-groups of N is greater than 1.

3. The intersection of two distinct Sylow p-groups of N is equal to

4. Every Sylow p-group of N is the intersection of N with exactly one Sylow p-group of .

5. The intersection of N with a Sylow p-group of which contains is a Sylow p-group of N.

6. The normalizer of a Sylow p-group of N in N is equal to the intersection of N with the normalizer of a Sylow p-group of which contains .

Proof: is in a Sylow p-group of and by hypothesis . Therefore by Theorem 6: . is a p-group in N, thus by Theorem 4 it lies in a Sylow p-group of N. By Theorem 4, lies in a Sylow p-group of . Since contains and thus is larger than , = . Therefore = ∩ N = is a Sylow p-group of N, and the in = N is uniquely determined by . Since every p-group in N is in a Sylow p-group of , the intersection of two distinct Sylow p-groups of N is equal to . Since is the intersection of two different Sylow p-groups of , N contains several Sylow p-groups. = N is a normal subgroup of n = N N. If we have

for an x in N, then it follows that , and therefore by 4., , consequently n_p = N N is the normalizer of p in N, Q.E.D.

As an application of this theorem, we shall show that every group of order pⁿq is solvable (p, q are two distinct primes).

If a Sylow p-group is a normal subgroup, then / is cyclic and by Theorem 6, Corollary 3, is solvable. Hence is solvable. Now suppose is not a normal subgroup of ; then : N = q, N = .

If the intersection of any two different Sylow p-groups is 1, then there are 1 + q. (pⁿ -1) elements of p-power order, and therefore there is at most one subgroup with order q. Consequently a Sylow q-group , is a normal subgroup of , and / is isomorphic to . Since /, and are solvable, is also solvable. Finally let be a maximal intersection of different Sylow p-groups greater than 1. The number of Sylow p-groups of N is > 1, is not divisible by p, is a divisor of pⁿq and therefore is equal to q. Also it follows from the previous theorem that lies in q different Sylow p-groups of . Therefore is the intersection of all the Sylow p-groups of . is a normal subgroup of , and the factor group / has as the maximal intersection of different Sylow p-groups the element 1. By what has already been proven, / is solvable. Moreover the p-group is solvable. Consequently is solvable, Q.E.D.

For many applications the following theorem is useful:

THEOREM 8 (Burnside): If the p-group in the finite group is a normal subgroup of one Sylow p-group but is not a normal subgroup of another Sylow p-group, then there is a number r, relatively prime to p, of subgroups conjugate to which are all normal subgroups of but which are not all normal subgroups of the same Sylow p-group of , so that the normalizer of transforms the _i transitively among themselves.

Proof: Among the Sylow p-groups which contain as a non-normal subgroup, , is chosen so that the intersection of , with the normalizer Nof is as large as possible. Let = ₁, ₂, …, _s be the subgroups conjugate to in the normalizer N of . Along with , all the _i, are also normal subgroups of . The normalizer N of contains N. Let ₁, ₂, …, _s, …, _r be all the groups conjugate to in N. Along with , all the _i are normal subgroups of . is contained in a Sylow p-group * of N N. Since is not a Sylow p-group of , while, by hypothesis, a Sylow p-group of N is also a Sylow p-group of , then * is larger than . * is in a Sylow p-group of N N, is in a Sylow p group of N and in a Sylow p-group of of . Since the intersection of with N contains *, and therefore is larger than , then by the construction of the Sylow p-group of is contained in N, and therefore a fortiori is contained in N.

Since contains the Sylow p-group of N N, we have = . Since therefore a Sylow p-group of N N is already a Sylow p-group of is relatively prime to p. If all the _i were

normal subgroups of the same Sylow p-group of , then the latter would be contained in N. But then the groups _i conjugate to each other in N would be normal subgroups in all the Sylow p-groups of N. Then would be a normal subgroup of the Sylow p-group ^* of the intersection of N with . But ^* is larger than , and this contradicts the definition of , as the intersection of with N; therefore the _i are not all normal subgroups of the same Sylow p-group of , Q.E.D.

The positional relationships of the subgroups of constructed in this proof can be seen from the diagram on the left.

§ 3. On p-groups

1. Nilpotent Groups.

Fundamental for the theory of p-groups is the following statement:

THEOREM 9: The center of a p-group different from e is itself different from e.

Proof: From the class equation for a group of order pⁿ > 1:

where the summands pⁱ run through indices > 1 of certain normalizers. Therefore : 1 is divisible by p, and consequently e.

COROLLARY:The (n + 1)-th member of the ascending central series of a group of order pⁿ is equal to the whole group.

The members of the ascending central series are defined as the normal subgroups _i of such that is the center of /_i. Now either _i = or, as just proven, _{i + 1} is larger than _i, and therefore certainly _n = .

By refinement of the ascending central series of a p-group we obtain a principal series in which every factor is of order p. It follows from the Jordan - Hölder - Schreier theorem that:

Every principal series of a p-group has steps of prime order.

The index of the center of a non-abelian p-group is divisible by p². This follows from the useful lemma: If a normal subgroup of a group is contained in the center and has a cyclic factor group, then is abelian. Since / is generated by a coset A, all the elements of are of the form AⁱZ where Z is in the center.

Therefore

and is abelian

If we apply the result found above to a p-group in which then: and since is abelian, it follows that:

The factor commutator group of a non-abelian p-group has an order divisible by p².

A group of order p or p² is abelian. In a non-abelian group of order p³, the center and the commutator group are identical and are of order p.

DEFINITION: A group is said to be nilpotent¹ if the ascending central series contains the whole group as a member, i.e., if

The uniquely determined number c is called, following Hall, the class of the group. Therefore “nilpotent of class 1” is the same as “abelian e.”

THEOREM 10: In a nilpotent group of class c it is possible to ascend to the whole group from any subgroup by forming normalizers at most c times.

Proof: Let be nilpotent of class c; let be a subgroup. Certainly ₀ is contained in . If _i is already contained in , then by the definition of _{i + 1}, it follows that _{i + 1} is contained in the normalizer of . By at most c repetitions of this procedure we obtain the result.

COROLLARY: Every maximal subgroup of a nilpotent group is a normal subgroup and therefore is of prime index.

Therefore in a p-group the intersection of all the normal subgroups of index p is equal to the Φ-subgroup defined earlier. The factor group /Φ is an abelian group of exponent p. By its order p^d the important invariant d = d() is defined. The significance of d is made clear by the following BURNSIDE BASIS THEOREM: From every system of generators of exactly d can be selected so that these alone generate . By the general basis theorem this theorem need only be proven for /Φ.

2. Elementary Abelian Groups.

An abelian group with prime exponent p is called an elementary abelian group. If is of order p^d then it is possible to generate by d elements: Let S₁ be an element e in : let S₂ be an element not in (S₁); let S₃ be an element of not in {S₁, S₂}; let S_d′ be an element of not in {S₁, S₂, …, S_d′-1} and {S₁, S₂, …, S_d′-1} = . Then (S_i) is of prime order p so that we must have

It follows from this that

and since : 1 = p^d, , we see that d = d¹. Therefore a finite elementary abelian group is the direct product of a finite number of cyclic groups of prime order. Conversely a direct product of a finite number of cyclic groups (S₁), (S₂), …, (S_d) of order p is an elementary abelian group. The elements S₁, S₂, …, S_d in the direct product representation are said to be a basis of . The above method of construction shows that every generating system of contains a basis. Therefore d is the minimal number of generators. Consequently every system of d generators is a basis of . The number of basis systems of can be calculated easily:

In the above construction there are p^d-1 possibilities for S₁; after choosing S₁, there are p^d-p possibilities for S₂ and so forth, so that we obtain the number (p^d – 1) (p^d – p)· … · (p^d – p^d−1) as the number of basis systems of . If S₁, S₂, …, S_d is a fixed basis and T₁, T₂, …, T_d is an arbitrary basis then the mapping

defines an automorphism of and conversely. Therefore it follows that:

The number of automorphisms of an elementary abelian group of order p^d is equal to . If we set

then the number is equal to . From the general basis theorem in II, § 4, it follows that:

The number of automorphisms of a p-group of order pⁿ (n > 0) and d generators is a divisor of

Remark: The highest power of p which divides this number is , and since 0 < d n, this number is a divisor of , as can easily be seen. Therefore the number of automorphisms of an arbitrary group of order pⁿ is a divisor of the number of automorphisms of the elementary abelian group of order pⁿ.

For later theorems it is important to obtain several formulae about the number φ_{q, α} of subgroups of order p^α in the elementary abelian group of order p^d. Let 0 < α d. Every subgroup of order p^α is elementary.

If S₁, S₂, …, S_a are the first α elements of a basis of the whole group, then these α elements are a basis of a subgroup of order p^α. Conversely, as we have seen previously, every basis of a subgroup of order p^α can be extended to a basis of the whole group. Since the elements S₁, S₂, …, S_α can be chosen in

different ways, and every subgroup of order p_α has different basis systems, then

(1)  

where k₀ = 1. From this the reader can derive the recursion formula

(2)  

for , where If we set for rational integers α which are larger than d or smaller than 0, then the formula is valid generally. From this formula we derive the congruence

(3)  

and the polynomial identity

(4)  

by induction. If we set x = 1, then

(5)  

An abelian group of order pⁿ can be decomposed, by the basis theorem, into the direct product of cyclic groups of orders p^n₁, p^n₂, …, p^n_r. Here the exponents n₁, n₂, …, n_r are determined uniquely to within order. Therefore we say: The group is of type (p^n₁, p^n₂, …, p^n_r). If we order the by size so that p^j occurs a_j times as the order of a basis element, then we say: The group is of type

Here the non-negative integers α_i are bound only by the relation

3. Finite Nilpotent Groups.

The direct product of a finite number of nilpotent groups is nilpotent, as is easily seen. For example, the direct product of a finite number of p-groups is nilpotent. The following converse is important:

THEOREM 11: Every finite nilpotent group is the direct product of its Sylow groups.

Proof: The normalizer of a Sylow group is its own normalizer by Theorem 5, and therefore, by Theorem 10, it is equal to the whole group; consequently every Sylow group is a normal subgroup.

Let p₁, p₂, …, p_r be the various prime divisors of the group order, and assume we have already shown that

Then the normal subgroups S_p1 · S_p2 · … · S_pi and S_Pi+1 have relatively prime orders so that their intersection is e; and therefore

But from the equation it follows, by comparing the orders, that the whole group is the direct product of its Sylow groups.

THEOREM 12: The Φ-subgroup of a nilpotent group contains the commutator group.

Proof: As we saw earlier, the Φ-subgroup is equal to the intersection of the whole group with its maximal subgroups. By the Corollary to Theorem 10, every maximal subgroup of a nilpotent group is a normal subgroup of prime index, and therefore every maximal subgroup of a nilpotent group contains the commutator group. Consequently the Φ-subgroup of a nilpotent group contains the commutator group.

Remark: We have further that , which can be derived from the definition of the Φ-subgroup as the intersection of the whole group with its maximal subgroups.

For finite groups we have the converse:

THEOREM 13 (Wieland): If the Φ-subgroup of a finite group contains the commutator group, then the group is nilpotent.

Proof: As in the proof of Theorem 11 it suffices to prove that every Sylow group is a normal subgroup. If the normalizer of a Sylow group were not the whole group, then it would be contained in a maximal subgroup which on the one hand would contain the Φ-subgroup and therefore the commutator group; and on the other hand, by Theorem 5, must be its own normalizer. Since this is not possible, every Sylow group must be a normal subgroup of the whole group.

THEOREM 14 (Hall): If the normal subgroup is not contained in _i but is contained in _{i + 1}, then the following is a normal subgroup chain without repetitions:

Proof: We have . Since is not contained in _i, (, ) is not contained in _i–1, and therefore is not contained in . We apply the same argument to , etc., Q.E.D.

  4. Maximal abelian normal subgroups.

It is natural to consider the maximal abelian normal subgroups as well as the maximal abelian factor group. In general abelian normal subgroups which are contained in no other abelian normal subgroup are neither uniquely determined nor isomorphic to each other, as is easily seen in the example of the dihedral group of eight elements. The center seems to be more appropriate as a counterpart of the factor commutator group, as we already have seen in the theorems on direct products.

In any case, there is, in every group whose elements e = a₁, a₂, …, are well ordered, a maximal abelian normal subgroup. We can construct an abelian normal subgroup _ω for any index ω in the following way: ₁ = e; let _ω be the union of all _r with v < ω; let _ω be equal to the normal subgroup generated by _ω and a_ω if this normal subgroup is abelian. Otherwise let = _ω = ω. The union of all the _ω is a maximal abelian normal subgroup.

A maximal abelian normal subgroup of a nilpotent group is its own centralizer.

Proof: The centralizer is a normal subgroup of . If contained properly, then by Theorem 14, a center element X in / would be contained in / 1 so that the subgroup generated by X and would be larger than . But since this subgroup containing would also be an abelian normal subgroup, we must have = .

If and are of orders pⁿ and p^m, respectively then the index p^n–m is a divisor of the number of automorphisms of , whereupon, by Part 2, it follows that

  5. The automorphism group of Z_N.

We wish to determine the automorphism group of the cyclic group Z_N for N > 1. For this purpose we consider Z_N as the residue class module (quotient module) of the additive group of integers with respect to the submodule of integers divisible by N. The operators of Z_N are given by the multiplications by the rational integers t; and are equal if and only if t₁ and t₂ are congruent mod N. is an automorphism if and only if t is relatively prime to N. The number φ(N) of automorphisms of Z_N is equal to the number of residue classes (cosets) mod N which contain numbers relatively prime to N (prime residue classes).

The automorphism group of Z_N (cyclic group of order N) is isomorphic to the group of prime residue classes mod N. If N is the product of relatively prime numbers m₁, m₂, then Z_N is the direct product of two characteristic cyclic groups of orders m₁, m₂. For the automorphism group we have the corresponding situation; in particular

If N is the n-th power of a prime p then a residue class is prime if and only if it consists of numbers relatively prime to p; the number of these residue classes is pⁿ – p^{n – 1}. If is the prime power decomposition, then

The residue class ring is a field and therefore, by II, § 7, the automorphism group of Z_p is cyclic of order p–1. A rational number g whose order mod p is p–1 is said to be a primitive congruence root mod p. g has an order which is divisible by p–1 mod pⁿ; say therefore, it has the order (p — 1) · p^r. The order of is then equal to p-1, mod pⁿ. If a = 1 + kp^m, then it follows from the binomial theorem that . Therefore a ≡ l(p^m) implies that . However if m > 1 or if p is odd then implies that . If p is odd, then 1 + p is of order p^n–1 mod pⁿ, (1 + p).g₁ is of order . If p = 2, then 1 + 2² is of order 2^n–2 mod 2ⁿ (n > 2). Since –1 is congruent to no power of 5 mod 4, there are, mod 2ⁿ, the 2^n–1 different prime residue classes . As a result we obtain:

If n < 3 or p is odd, then the automorphism group of is cyclic of order (p — l)p^n–1. The automorphism group of Z_2n, for n > 2, is abelian of type (2^n–2, 2) with the associated basis automorphisms 5 and –1.

6. p-Groups with only one Subgroup of Order p.

A non-cyclic abelian group of exponent pⁿ contains at least two different subgroups of order p.

Proof: Let A be an element of order pⁿ and let B not be a power of A. Then the order p^r of B mod (A) is greater than 1, but at most pⁿ. We have

Therefore and generate two different subgroups of order p.

We wish to find non-abelian groups of order pⁿ which contain only one subgroup of order p.

An example is the quaternion group. By the theorem of Hölder it is defined by the relations as a group of order 8 with generators A and B. Its eight elements are called quaternions; they are

If instead we write

and set

then we have the following calculational rules:

From this we conclude that there is only one subgroup of order 2 and exactly three subgroups of order 4. The center is equal to the commutator group which is equal to (—1).

The generalized quaternion group is defined by the relations

as a group generated by A, B, and of order 2ⁿ, by the Hölder Theorem. Since

this group contains only one subgroup of order 2. The elements and B generate a quaternion group.

The relations above can be written more elegantly in the form

The new relations follow from those above.

From the new relations, however, it follows that

and therefore the old relations follow.

If A′ is of order 2^n–1, B′ of order 4, and if A′ and B′ generate the whole group, then

Therefore all the calculational rules which are valid for power products of A and B also remain valid for the corresponding power products of A′ and B′

Since A′ and B′ generate the whole group, (A′) is a normal subgroup of index 2, and every element can be written uniquely in the form

Therefore the mapping is an automorphism of the group. The number of all the automorphisms is equal to the number of pairs A′, B′ It follows by simple enumeration that:

The quaternion group has exactly 24 automorphisms. The generalized quaternion group of order 2ⁿ has exactly 2^2n–3 automorphisms for n > 3.

In the automorphism group A of the quaternion group, the inner automorphisms form an abelian normal subgroup J of order 4. An automorphism which commutes with all the inner automorphisms is itself an inner automorphism. Since it changes each generator by a factor in the center, there are at most 2·2 such automorphisms.

A group A having order 24, and containing a normal subgroup J of order 4 which is its own centralizer, must be isomorphic to ₄.

This is because a central element of A must be in J, and an element of order 3 must transform the three elements e in J in a cyclic manner. Then, since according to the results of Sylow there are elements of order 3, the center is e, and there is no normal subgroup of order 3. From these results also, the index in A of the normalizer N₃ of a Sylow 3-group is 4. A transitive representation of A in 4 letters is associated with N₃. The representation is faithful since the intersection of all 3-normalizers contains only center elements with orders 1 or 2 and therefore is e. Since A consists of 24 elements, A is isomorphic to ₄.

The automorphism group of the quaternion group is isomorphic to the symmetric permutation group of four letters.

The quaternion group is the only p-group tvhich contains two different cyclic subgroups of index p but only one subgroup of order p.

Proof:¹ Let be of order pⁿ and let it contain two different cyclic subgroups ₁ and ₂, of index p. ₁ and ₂ are different normal subgroups of index p, and therefore their intersection is of index p². Moreover is in the center and contains the commutator group. It follows for any two elements x, y that x^p and y^p are in , and that

If p is odd, then (xy)^p = x^py^p, and therefore the operation of raising to power p is a homomorphy. Since the group of p-th powers is contained in , by the first isomorphism theorem the elements whose p-th power is e form a subgroup whose order is at least p² There are at least two different subgroups of order p in this subgroup.

If p = 2, then (xy)⁴ = (y, x)⁴ x⁴y⁴ = x⁴y⁴. Now we conclude just as above that either = 1 and ₁, ₂ are two different subgroups of order 2, or there are two subgroups _{1 2} of order 4 by the first isomorphism theorem. We may assume that ₁ is in . If ₁ is different from ₁, then ₁ is in and ₁ · ₂ is an abelian group of order 8. Since it contains two different subgroups of index 2, it is not cyclic, and therefore it also contains two different subgroups of order 2. If, in conclusion, ₁ = ₁, then the whole group is of order 8. Let = (A) and ₂ = (B). If there is only one subgroup of order 2 then B² = (A B)² = A², and therefore the group is the quaternion group.

THEOREM 15: A p-group which contains only one subgroup of order p is either cyclic or a generalized quaternion group.

Proof: Let be of order pⁿ and let it contain only one subgroup of order p. First let p be odd. If n = 0, 1, then the theorem is clearly true. We now apply induction to n.

Every subgroup of index p is cyclic by the induction hypothesis, and therefore by what was proven previously there is only one subgroup of index p in and , therefore itself is cyclic.¹

Now let p = 2 and let be a maximal abelian normal subgroup. is cyclic and its own centralizer. Therefore / is isomorphic to a group of automorphisms of . We shall show that only one automorphism of order 2 can occur, namely, the operation of a raising the elements of to the power –1. Since this automorphism is not the square of any other automorphism of , it follows that : is either 1 or 2. If we set = (A) and assume that B e(), B² ≡ e(), then as a preliminary BAB^–1 must be shown to be equal to A^–1. In fact, we want to show further that the group generated by A and B is a generalized quaternion group with relations (8) and (9). Then the theorem will be proven.

Since B cannot commute with all the elements of A, (B²) A, and there is a subgroup ₁ of which contains (B²) as a subgroup of index 2. The group ₁(B) contains the two different cyclic subgroups ₁ and (B) of index 2; and therefore it is, as was previously shown, the quaternion group. If A is of order 2^m then: B² = A^{2m – 1}. We also conclude (AB)² = A^{2m – 1} Therefore A and B generate the generalized quaternion group of order 2^{m + 1}.

THEOREM 16: A group of order pⁿ is cyclic if it contains only one subgroup of order p^m (where 1 < m < n).

Proof: There is a subgroup of order p^m. is contained in a subgroup ₁ of order p^{m + 1} and is the only subgroup of index p in ₁. Therefore ₁ is cyclic and consequently is cyclic. Since every subgroup of order p or p² is contained in a subgroup of order p^m, and since the only subgroup of order p^m is cyclic, there is only one subgroup of order p and one of order p². Since the generalized quaternion group contains some subgroups of order 4, we conclude from the previous theorem that the whole group is cyclic.

If in a p-group, every subgroup of order p² is cyclic, then there is only one subgroup of order p, and conversely.

If there were two different subgroups of order p then we can assume that one of them is contained in the center. But then the product of the two subgroups is a non-cyclic group of order p². Conversely, in a non-cyclic group of order p² there are certainly two different subgroups of order p. Now one can easily prove:

THEOREM 17: A group of order pⁿ in which every subgroup of order p^m is cyclic, where 1 < m < n, is cyclic except in the case p = 2, m = 2 in which case the group can also be a generalized quaternion group.

7. p-Groups with a Cyclic Normal Subgroup of Index p.

We shall determine all the p-groups which contain a cyclic normal subgroup of index p. This problem will now be solved for non-abelian p-groups, which contain some subgroups of order p. If is of order pⁿ then in there is an element A of order p^{n – 1} and an element B of order p which is not a power of A. A and B generate , and the subgroup (A) of index p is a normal subgroup. Therefore

If for odd p the element B is replaced by an appropriate power, then we can take r = 1 + p^{n – 2}.

If p = 2, n = 3, then we must have r ≡ — 1 (4). If p = 2, n > 3, then there are three possibilities for r,

The number r is not altered mod 2^{n – 1} if B is replaced by BA^μ.

If then the commutator subgroup is of order 2; in the other two cases it is of order 2^{n –1}.

If r ≡ — 1, then and therefore there is only one cyclic subgroup of index 2. Thus r is uniquely determined by the group. As a result we obtain:

The groups of order p^{u – 1} which contain an element A of order p^{u –1}, are of the following types:

a) abelian:

b) non abelian, p odd:

c) non-abelian, p = 2:

Groups of different type are not isomorphic. From Hölder’s theorem it follows that all types exist. For n = 3, V will coincide with IV, and VI with II.

Now it is simple to give all groups of order p³. We must now investigate among such all those in which the p-th power of every element is e. A group in which all squares are equal to e is abelian since

If the group is non-abelian and p is odd, then it is generated by two elements A and B such that the relations

hold. By III, Theorem 21, these relations define a non-abelian group with generators A, B and order p³, in which, for any two elements x, y, we have:

Thus the p-th power of every element is equal to e. As a result we obtain:

There are p for every prime number p, five types of groups of order p³, namely the three abelian types:

and two non-abelian types, which are, for p = 2; III, quaternion group and IV, the dihedral group, and for odd p the types

Exercises

1. If a p-group contains a cyclic normal subgroup of index p, then every subgroup different from e has the same property.

2. For odd p, the following properties hold for abelian groups of type (p, p^{n – 1}) and for non-abelian groups of order pⁿ having a cyclic subgroup of index p, where m is a number greater than zero and less than n:

a) The number of subgroups of order p^m is 1 + p in both cases.

b) The number of cyclic subgroups of order p^m is, in both cases, 1 + p or p according to whether m = 1 or m > 1.

c) The number of elements whose p^m-th power = e is p^{m + 1} in both cases.

d) In both groups, every subgroup whose order is divisible by p² is a normal subgroup. Therefore for m > 1 there are equally many normal subgroups of order p^m.

e) The number of automorphisms is pⁿ(p-1).

3. The two types of non-abelian groups of order p³ can be defined by the relations

for all p by an appropriate choice of generators A, B.

4. If a 2-group contains a cyclic subgroup of index 2 and is neither abelian of type (2, 2) nor the quaternion group, then the number of its automorphisms is a power of 2.

5. In a finite group, the index of the normalizer over the centralizer of a Sylow p-group with d generators is a divisor of k_d. If the order of the group is divisible neither by the third power of its smallest prime factor p, nor by 12, then every Sylow p-group is in the center of its normalizer.

6. In an abelian p-group with the exponent p^m, the characteristic chains

and  

give rise to a characteristic series through the refinement process which was given in the proof of the Jordan-Hölder-Schreier Theorem. There is only this one characteristic series. (Here ^pv denotes the group of the p^r-th powers and _p^v denotes the group of all elements whose p^v-th power is e.)

7. Theorem 2 in § 1 admits the following corollaries: If is a Sylow p-group, in , N its normalizer, a normal subgroup of , then

a) N / is the normalizer of the Sylow p-group of / of /;

b) N is contained in the normalizer N of the Sylow p-group = ∩ of ;

c) N = ; therefore by the Second Isomorphism Theorem

(Hint for a): If xx^–1 = , then by Theorem 3: xx^–1 = vv^–1 is solvable for v in , therefore v^–1x : for every x in , xx^–1 = vv^–1 is solvable for v in .)

With the help of c) it should be shown that the φ-subgroup of a finite group is nilpotent.

§ 4. On the Enumeration Theorems of the Theory of p-Groups

In the study of finite groups the question arises naturally as to the number of elements or subgroups with some given property. The results obtained in connection with this question do not lie very deep.

The following systematic derivation of the enumeration theorems in p-groups is due to P. Hall.

THEOREM 18: (Counting Principle): Let be a finite p-group. _α denotes any subgroup of index p^α which contains Φ(). Let () be a set of complexes such that each complex in () is contained in at least one subgroup of index p. Let n(_α) be the number of complexes of () which are contained in _α. Then

where the summation is extended over the φ_{d, α} subgroups _α of .

Proof: We shall show that the number of times that an element in () is “counted” with the appropriate sign on the left of the equation above is equal to zero.

The intersection of all _α which contain , contains φ() and therefore is an . By hypothesis is contained in an ₁ and therefore > 0. The number of all _α’s which contain is equal to the number of all _α’s which contain , i.e., φ_{, α.} Therefore the number of times that is “counted” is

But this number is zero, by §3, Formula 5, Q.E.D.

THEOREM 19: The number of subgroups of fixed order p^m(0 ≤ m ≤ n) of a p-group of order pⁿ leaves 1 as a remainder when divided by p.

Proof: If n = 0, then the theorem is clear. Now let n > 0 and assume that the theorem is proven for p-groups whose order is less than pⁿ. If m = n then the theorem is trivial. Let m < n. For Theorem 19, let () denote the set of all subgroups of of order p^m. Then:

and by the induction hypothesis

moreover the number of all ₁ is φ_{d, d –1}, therefore by §3 congruent to 1 mod p, so that

follows, Q.E.D.

THEOREM 20: (Kulakoff): In a non-cyclic p-group of odd order pⁿ, the number of subgroups of order p(0 < m < n) is congruent to 1 + p modulo p².

In the non-cyclic group of order p² there are p + 1 subgroups of order p. We apply induction on n and assume n > 2.

The number of all ₁ is φ_{d, d – 1}, and therefore, since d > 1, is congruent to 1 + p mod p². Let m < n-1, () be the set of all subgroups of order p^m. By the Counting Principle it follows that

By Theorem 19

and by §3, 2., the number of all ₂ (namely φ_{d, d – 2}) is congruent to 1 mod p. Consequently

For the non-cyclic by the induction hypothesis. As was shown, the number of all ₁ is congruent to 1 + p(p²). If there is no cyclic subgroup of index p in , then

If contains a cyclic subgroup of index p, then the theorem follows from the solution of Exercise 2a at the end of §3.

THEOREM 21 (Miller): In a non-cyclic group of odd order pⁿ, the number of cyclic subgroups of order p^m (1 < m < n) is divisible by p.

Proof: If contains a cyclic subgroup of index p, then the theorem follows from the solution of Exercise 2b at the end of §3. To continue, let every subgroup of index p in be non-cyclic, m < n-1 and assume the proof has been carried out already for smaller n. Let () be the set of cyclic subgroups of order p^m. We find the congruence:

By the induction hypothesis each of the numbers n(₁) is divisible by p, and therefore the desired number n(₀) is also.

THEOREM 22 (Hall): The number of subgroups of index p^α in is congruent to φ_{d, d}(mod p^{d – α + 1}). The number of those subgroups which do not contain Φ() is consequently divisible by p^{d – α + 1}.

Proof: If d = n, then the number in question is already known to be φ_{d, α.} Let n > 1 and let the theorem be proved for smaller n. If α = 0, then the theorem is clearly true. Let α = 0, then n(_β) is equal to the number of all subgroups of index p^{α – β} in _β. Therefore n(_β) = 0 if β > α; but otherwise by the induction hypothesis

Since and therefore by § 3

we have

The Counting Principle now gives the congruence

But by the Counting Principle, the right side of the congruence is exactly the number of subgroups of index p^α in an elementary abelian group of order p^d, so that

Exercise (Kulakoff): In a non-cyclic p-group of odd order pⁿ, the number of solutions of x^pm = e(0 < m < n) is divisible by p^{m + 1}.

Exercise: The number of normal subgroups of order p^m in a group of order pⁿ(0 < m < n) is congruent to 1 (mod p).

If p is odd, 1 < m, and is non-cyclic then, more precisely, the number is congruent to 1 + p (mod p²).

§ 5. On the Descending Central Series

P. Hall has generalized the concept of a terminating ascending central series by defining:

A chain of normal subgroups of

is called a central chain if _i/_{i + 1} is contained in the center of /_{i + 1} (i = 1, 2, … r).

If the ascending central series (See II § 4, 3.) terminates, then it is a central chain. The following definition is still more useful: A chain of subgroups

is said to be a central chain if the mutual commutator group (, _i) is contained in _{i + 1} (i = 1, …, r). Since for every x_i in _i, x in : and thus certainly xx_ix^–1 _i, it follows that _i is a normal subgroup of and that _i/_{i + 1} is contained in the center of /_{i + 1.} The converse is clear. _{r + 1} is contained in ₀; if it has already been shown that _{r + 1 – i} is in _i where i < r, then

and therefore . Hence

Consequently = .

If a group has a central chain, then it is nilpotent and the length of every central chain is at least equal to the class of the group.

Now it is natural to define the descending central series for an arbitrary group as where

If has a central chain (1) then it follows by induction that: and therefore If, conversely, the descending central series is equal to e from the (r + 1)-th place on, then is a central chain. If c is the class of , then and therefore

In a nilpotent group, the class c can be found from the relation:

By Chapter II, § 6, _i is a commutator form of of weight¹ i and of degree 1 and is generated by the higher commutators (G₁ G₂, … G_i) where G_j . Therefore _i is a fully invariant subgroup of .

For every subgroup of it follows that

If is a normal subgroup of , then

It follows from this that:

Every subgroup and every factor group of a nilpotent group is itself nilpotent, and the class of the subgroup or factor group is at most equal to the class of the whole group.

We wish to state something about the positional relationships, and the mutual commutator groups, of members of the descending central series and of an arbitrary central chain.

If is a sequence of subgroups of an arbitrary group , so that it follows immediately that is a normal subgroup of . If moreover then it follows by induction that In nilpotent groups of class c we can conclude from this that:

(4)  _i is contained in _{c – i} (since otherwise we would have _c = e).

1 _i is also called the i-th Reidemeister commutator group.

Now we claim that in the general case

We carry out the proof by induction on i. By hypothesis

Let i > 1 and assume we have already proven that for all k.

Then by II. Theorem 14:

and by the induction hypothesis:

Therefore

If we set

then

We can now show by induction on the weight that:

An arbitrary commutator form f() of weight w is contained in _w.

This is true if w = 1. Now let w > 1, and assume that the statement is already known to be true for commutator forms with weight less than w. We have f() = (f₁(), f₂()) where the f_i are commutator forms of weight such that w = w₁ + w₂. By the induction hypothesis it follows that thus

In particular it follows that

A nilpotent group of class c is always k-step metabelian, where k satisfies the inequality

Moreover if we set _–1 = _–2 = … = e then in general

In particular, _i, commutes with _i elementwise.

THEOREM 23 (Hall): If the non-abelian normal subgroup of the p-group is contained in _i, then its center is of order at least pⁱ itself is at least of order p^{i + 2}, its factor commutator group is at least of order p^{i + 1}.

Proof: Since _i commutes with _i elementwise, is not contained in is in the center of and, by Theorem 14, is at least of order pⁱ so that a fortiori the center of is of order divisible by p^{i + 2}. Since p²/: (), the order of is divisible by p^{i + 2}. Since is not abelian, we can find in the normal subgroup ′ of a normal subgroup ₁ of with , of index p under ′. /₁ is a non-abelian normal subgroup of /₁ and so we conclude as above that :₁ is divisible by p^{i + 2}. Consequently /′ has an order divisible by p^{i + 1}, Q.E.D.

Now if in a p-group of order pⁿ, then and therefore as was just shown, . If is now (k + 1)-step metabelian, then

The order of a (k + 1) step metabelian p-group is divisible by p^{2k + k}.

Remark: Under the hypothesis of Theorem 23 it can be shown by the same methods that the factor groups of the ascending and descending central series of the normal subgroup have an order divisible by pⁱ, with the possible exception of the last factors different from 1. The proof is left to the reader.

Exercises

1. In a finite group the intersection of all the normal subgroups whose factor group is an abelian p-group is called the p-commutator group of and is denoted by ′(p).

Prove: The p-factor commutator group /′(p) is an abelian p-group. Moreover, the commutator group of is the intersection of the p-commutator groups, and the factor commutator group is isomorphic to the direct product of the p-factor commutator groups. Moreover,

2. For an arbitrary group , the class may be defined by the following property: Let the class be equal to c, if _c+1 is a proper subgroup _e and _{e + 1} = _{e + 2} = … Let the class be equal to zero, if the group coincides with its commutator group.¹ Let the class be infinite if _{i + 1} is a proper subgroup of _i for all i.

For nilpotent groups the two definitions of class coincide.

Prove: If the class c is finite then _{e + 1} is the intersection of all normal subgroups with nilpotent factor group, and the factor group /_{e + 4} is also nilpotent. Hence we shall call the factor group /_{e + 1} the maximal nilpotent factor group. Its class is c . The class of every factor group is at most c.

If the class is infinite, then there are factor groups of any given class.

3. In finite groups of class c we can obtain _{e + 1} the following way:

For every prime number p we form the intersection _p() of all normal subgroups of p-power index.

Prove: _p itself is of p-power index. Hence we shall call the factor group / _p the maximal p-factor group of .

Prove: _e+1 is the intersection of all and the maximal nilpotent factor group is isomorphic to the direct product of the maximal p-factor groups over all prime divisors of the group order.

4. If p^a is divisible by the exponent of the maximal p-factor group of the finite group (see Exercise 3), then the subgroup generated by all p_a-th powers is equal to _p. Therefore _p is a fully invariant subgroup of . Moreover, prove that

5. a) An abelian group with a finite number of generators is finite if and only if the factor group over its Φ-subgroup is finite.

b) A nilpotent group with a finite number of generators is finite if and only if the factor group over its Φ-subgroup is finite. [Use a) and apply induction to the length of the descending central series!]

6. In a nilpotent group all the elements of finite order form a fully invariant subgroup. (Use Exercise 5.)

7. a) (Hilton.) In a nilpotent group any two elements with relatively prime orders commute.

(Hint: Show that the commutator of the two elements is in members of the descending central series with arbitrarily great subscript.)

b) Two elements with p-power order generate a p-group.

c) Prove the following generalization of Theorem 11: A nilpotent group in which every element is of finite order is the direct product of nilpotent groups in which every element is of prime power order.

§ 6. Hamiltonian Croups

In an abelian group every subgroup is a normal subgroup. What other groups also have this property?

DEFINITION: A non-abelian group in which every subgroup is a normal subgroup is said to be a Hamiltonian group. For example, the quaternion group is a Hamiltonian group.

THEOREM: A Hamiltonian group is the direct product of a quaternion group with an abelian group in which every element is of odd order and an abelian group of exponent 2, and conversely.

Proof: In a Hamiltonian group there are two elements A, B which do not commute with each other. Since (A) and (B) are normal subgroups of , the commutator of A and B is contained in the intersection of (A) and (B), and therefore in the center of the subgroup = {A, B} generated by A and B.

The commutator group of ' of is generated by C and is a proper subgroup of (A) and likewise of (B). Since where r, s 0. By Chapter II § 6, (A, B)^s = (A, B^s), and therefore C^s = e. Consequently A and B have finite orders m and n respectively. We choose A and B so that m and n are minimal. Then it follows for a prime divisor p of m that

Similarly it follows for a prime divisor p of n that C^p = e. The orders of A, B are consequently powers of the same natural prime p ; they are divisible by p² since (C) is a proper subgroup of both (A) and (B), while A^p, B^P are contained in the center of .

If, say , where p, μ are not divisible by p, then we replace A by A_μ, B by B^v, and we may assume that

where a b > 0.

By chapter II § 6, in we have the relation

Now also generate , and therefore must be of order at least equal to that of B. From this we conclude:

Therefore is a quaternion group with the relations A² = B² = ABA^-1B^-1 = C, C² = e.¹

We wish to show that, is generated by and the group of all elements of which commute with every element of .

If the element X does not commute with A, then X A X^-1 = A^-1 and therefore BX commutes with A. If, now, BX does not commute with B, then ABX commutes with B. Consequently.

Every element X in is of finite order, since BX does not commute with A, and therefore BX is of finite order. But B is of order 4 and commutes with X, therefore X is of finite order. Now, if X⁴ = e, X ε, then (A, BX) e, (A,BX) = A² = B². Since (BX)⁴ = e, we have (A,BX) = (BX)² = B²X2 and therefore X² = e.

In there is no element of order 4 and thus certainly no quaternion group. But since every subgroup in is a normal subgroup, is abelian. is the direct product of the subgroup of all elements of odd order, and the subgroup , of all elements whose square is e. C is contained in . Among all the subgroups of , which do not contain C there is a largest . For every element X in , not contained in , C must be contained in { , X}. Since X² = e, we have {, X }: = 2 and likewise { ,C}: = 2, and therefore {, X } = {, C}; it follows that {, C\ = i and moreover (C) = e; therefore

Since we have and moreover therefore .

Conversely a group with this structure is Hamiltonian. For is not abelian. We have yet to show that every cyclic subgroup (QUG) is a normal subgroup. Since is the only non-abelian factor of the decomposition we only need show that the transform of QUG by A or B is in (QUG).

Now where i is either 1 or 3. The order of U is an odd number d. Therefore the congruences can be solved,and

§ 7. Applications of Extension Theory

Let be an extension of the normal subgroup with the factor group

We say a factor system (C_aτ) is an abelian factor system if all the C_στ, commute with each other.

THEOREM 24: The (:1)-th power of an abelian factor system is a retracting¹ factor system.

Proof: Let (: 1) = n > 0. We set and form the product over of all the equations .

Then it follows that

THEOREM 25 (Schur): If the order n of the finite factor group is relatively prime to the order m of the finite normal subgroup , then the extension splits over .

Proof: We need only show that contains a subgroup of order n.

If m = 1, this is clear. Let m > 1 and assume the statement proven when the order of the normal subgroup is less than m. For a prime divisor p of to, every Sylow p-group S_p of is contained in . Since there are as many Sylow p-groups in as in , N_p: N_p Now N_p /S_p, is a normal subgroup of N_p/S_p with index n. By the induction hypothesis there is a subgroup /S_p of order n in N_p/S_p. S_p/z_p is a normal subgroup of /z_p of index n, where z_p is the center of S_p and is different from e. By the induction hypothesis there is a subgroup /S_p of order n in /z_p. Let C_{σ τ} be a factor system of over z_p. Since the order z of z_p is relatively prime to n, we can solve the congruence nn_L ≡ 1 (z) and for the factor system C_{σ, τ} of over z_p we find that it is the n₁-th power of the factor system which is retracting by Theorem 24. Therefore C_{σ, τ} itself splits over z_p, i.e. contains a subgroup of order n, Q.E.D.

In what follows, let be a finite group of order n.

THEOREM 26: If and the a_σ commute with one another, then the equation is solvable, i.e. the mapping can be accomplished by transformation with an element δ in .

Proof: Form the product over all equations with fixed σ:

We set and have

It has been conjectured that the following theorem is true in general.

THEOREM 27: If the order n of the finite factor group is relatively prime to the order m of the finite normal subgroup , then two representative groups of over are conjugate in . We shall prove the theorem when one of the folloiving additional conditions holds:

1. is abelian.

2. is solvable.

3. is solvable.

One of the groups , is of odd order, and since it is conjectured that groups of odd order are solvable, it is also expected that the above theorem is true. E. Witt reduced the theorem to the case when is simple and the centralizer of in is e. It is believed that the group of outer automorphisms of a finite simple group is solvable, so that we can conjecture the truth of Theorem 27 on this basis also.

Proof of 1: If = {S_σ} is a representative group and = {a_σS_σ} a second one, then we have the equations

By Theorem 26, is solvable. Since by hypothesis the congruence is solvable, , Q.E.D.

Proof of 2: If is abelian, then the theorem is true by 1; let be k-step metabelian and assume the theorem has already been proven for D^{k – 1}() = e. Further, let and be two representative groups of over . We apply 1 to /′ and find that ′ = (′)^x with x is solvable i.e., . Since , then by applying the induction hypothesis to ′, it follows that is also solvable for y and therefore = ^xv, with xy , Q.E.D.

Proof of 3: Let a principal series of / be of length l, and let , , be two representative groups of over . Let u be a minimal normal subgroup of ; since is solvable, u is a p-group. u is isomorphic to where is a normal subgroup of .

If l = 1, then u = , = . Then and are Sylow p-groups of therefore conjugate in . Let l > 1, and assume that the theorem has been proven for smaller l. By the induction hypothesis there is an x in u = , such that = u^x. We set ₁ = ^x–1 and find that , . Since the principal series of /u is of length l-1, it follows by the induction hypothesis applied to N_u/u that there is a y N_u, such that = ^v, and therefore = ^xv, Q.E.D.

Exercises

In a finite solvable group, certain generalized Sylow theorems are valid (Hall):

1. For every decomposition N = n.m of the group order into a product of relatively prime factors, there is a subgroup of order m and index n.

2. Let n and m be chosen as in Exercise 1. All subgroups whose order is a divisor of m lie in a subgroup of order m.

3. Letm be as in Exercise 1. All subgroups of order m are conjugate. The normalizer of a subgroup of order m is its own normalizer.

(Proofs of 1-3 by induction on the length of the principal series and by use of Theorems 25, 27.)

¹ According to a communication from E. Witt.

² If is the intersection of two Sylow p-groups and no group containing property is contained in the intersection of any two Sylow p-groups, is called a maximal intersection of two Sylow p-groups.

³ This name is used because for finite continuous groups the associated Lie ring of infinitesimal transformations is nilpotent precisely when the ascending central series terminates with the full group.

⁴ Here we must anticipate the result of § 5 which is trivial for p-groups; namely, that every factor group of a nilpotent is itself nilpotent.

⁵ In accordance with a communication from Herr Maass, Hambirg.

⁶ This last by the basis theorem.

⁷ is also called the i-th Reidemeister commutator group.

⁸ These groups are also said to be perfect groups.

⁹ Instead of this process one can apply Theorem 15 of § 3 to the group !

¹⁰ See end of § 6, Chapter III.