Exponent-critical groups

A finite group 𝐺 is exponent-critical if the exponent of 𝐺 is not the least common multiple of the exponents of its proper non-abelian subgroups.

The order of a non-abelian exponent-critical group is divisible by at most three distinct primes.

Let 𝐺 be a non-abelian finite group whose order is divisible by exactly three distinct primes. Then 𝐺 is exponent-critical if and only if 𝐺 is a direct product of a cyclic Sylow subgroup of 𝐺 and its complement, which is minimal non-abelian.

Let 𝐺 be a finite group, and let 𝑝 be a prime number. A subgroup 𝐻 of a group 𝐺 is a 𝑝-witness for 𝐺 if it is non-abelian, proper, and the 𝑝-parts of exp ⁡ ( G ) and exp ⁡ ( H ) are equal.

Let 𝐺 be a non-abelian exponent-critical group whose order is divisible by exactly two distinct primes 𝑝 and 𝑞. Without loss of generality, swapping 𝑝 and 𝑞 if necessary, we may suppose that 𝐺 does not possess a 𝑝-witness for 𝐺. Then 𝐺 is isomorphic to one of the following families of exponent-critical groups.

1. 𝐺 is a direct product of a cyclic Sylow 𝑝-subgroup of 𝐺 and its complement which is minimal non-abelian.

2. 𝐺 is a semi-direct product of a normal abelian Sylow 𝑝-subgroup 𝑃 of 𝐺 by a cyclic Sylow 𝑞-subgroup 𝑄. The subgroup 𝑃 is a direct product of cyclic groups of order p m for some positive integer 𝑚. We have | G : C G ( P ) | = q , and 𝑄 acts non-trivially and irreducibly on P / P p .

3. 𝐺 is a semi-direct product of a normal abelian Sylow 𝑝-subgroup 𝑃 of 𝐺 by a cyclic Sylow 𝑞-subgroup 𝑄. The subgroup 𝑃 is a direct product of a cyclic group of order p m and an elementary abelian group. We have m > 1 . The action of the subgroup 𝑄 on 𝑃 preserves this direct product, acting non-trivially and irreducibly on the elementary abelian group, and trivially on the cyclic group of order p m . We have | G : C G ( P ) | = q .

4. 𝐺 is a semi-direct product of a (unique) normal Sylow 𝑞-subgroup 𝑄 by a non-normal cyclic Sylow 𝑝-subgroup 𝑃. The subgroup 𝑄 is special in the sense of Gorenstein [8, p. 183], so 𝑄 is either elementary abelian or 𝑄 has nilpotency class 2 with Q ′ = Z ⁢ ( Q ) = Φ ⁢ ( Q ) and Q ′ is elementary abelian. Further, if x ∈ P generates 𝑃, then 𝑥 acts irreducibly on Q / Q ′ and trivially on Q ′ .

A non-abelian finite 𝑝-group has type ℬ if and only if it is 2-generated with derived subgroup of order 𝑝.

Let 𝑊 be the abelian group defined by

where a 0 has order p m and a i has order p m − 1 for 1 ≤ i ≤ p − 1 .

Let ϕ : W → W be the automorphism (see Lemma 4.5) of 𝑊 such that

Let ⟨ b 0 ⟩ be the cyclic group generated by an element b 0 of order p m − 1 . Define 𝑈 to be the semi-direct product U = W ⋊ ⟨ b 0 ⟩ , where b 0 acts on 𝑊 via the automorphism 𝜙. So b 0 − 1 ⁢ w ⁢ b 0 = ϕ ⁢ ( w ) for all w ∈ W .

Define D ⊴ U to be the normal subgroup of order 𝑝 generated by the element

Let 𝒩 be the set of normal subgroups 𝑁 of 𝑈 such that

The following statements hold.

1. Let 𝑃 be an exponent-critical 𝑝-group of exponent p m and type 𝒜. Then 𝑃 is isomorphic to U / N , where N ∈ N .

2. Any group of the form U / N with N ∈ N is a non-abelian exponent-critical 𝑝-group of exponent p m .

3. There are no exponent-critical 2-groups of exponent 2 2 and type 𝒜. When 𝑝 is odd or m ≥ 3 , the converse to (i) holds: any group of the form U / N with N ∈ N is an exponent-critical 𝑝-group of exponent p m and type 𝒜.

Proof of Theorem A

Let 𝐺 be a non-abelian exponent-critical finite group. We mentioned in the introduction that 𝐺 is solvable. Suppose that the order of 𝐺 is divisible by four or more primes. Since 𝐺 is non-abelian, it must be the case that some Sylow 𝑝-subgroup 𝑃 of 𝐺 is not central. Let 𝑞 be a prime dividing the order of 𝐺 such that C G ⁢ ( P ) does not contain a Sylow 𝑞-subgroup of 𝐺. Let H p ⁢ q be a p ⁢ q -Hall subgroup of 𝐺, which is necessarily non-abelian. For any other prime 𝑟 dividing | G | , a Hall p ⁢ q ⁢ r -subgroup H p ⁢ q ⁢ r of 𝐺 is therefore non-abelian. Moreover, H p ⁢ q ⁢ r is proper since | G | is divisible by four or more primes. So H p ⁢ q ⁢ r is a 𝑝-witness, a 𝑞-witness and an 𝑟-witness for 𝐺. So there exists an ℓ-witness for 𝐺 for all primes ℓ dividing the order of 𝐺, which contradicts 𝐺 being exponent-critical. ∎

Let 𝐺 be a non-abelian, exponent-critical group. If the order of 𝐺 is divisible by three distinct primes, then 𝐺 has a non-trivial central cyclic Sylow subgroup and all Sylow subgroups of 𝐺 are abelian.

Proof

By the same reasoning as in the proof of Theorem A, the group 𝐺 has a non-abelian p ⁢ q -Hall subgroup for two primes 𝑝 and 𝑞 dividing | G | . This subgroup is proper as three primes divide the order of 𝐺, and so it is a 𝑝-witness and 𝑞-witness for 𝐺. Suppose 𝑟 is the third prime dividing | G | . If either the p ⁢ r - or q ⁢ r -Hall subgroups are non-abelian, then we have an 𝑟-witness for 𝐺, and so 𝐺 is not exponent-critical. Therefore, both these Hall subgroups are abelian. Hence all Sylow subgroups of 𝐺 are abelian, and the Sylow 𝑟-subgroup centralizes a Sylow 𝑝- and Sylow 𝑞-subgroup of 𝐺. If the Sylow 𝑟-subgroup is not cyclic, then 𝐺 would contain a proper subgroup which is an 𝑟-witness for 𝐺, which would imply that 𝐺 is not exponent-critical. ∎

Proof of Theorem B

Suppose 𝐺 is exponent-critical. Then 𝐺 has a central cyclic Sylow 𝑝-subgroup 𝑃 by Theorem 2.1. Let 𝐻 be a Hall q ⁢ r -subgroup of 𝐺. Since 𝑃 is central, 𝐺 is a direct product of 𝑃 and 𝐻. If 𝐻 were abelian, then 𝐺 would be abelian. Hence 𝐻 is non-abelian. Suppose by way of contradiction that 𝐻 has a non-abelian proper subgroup 𝐾. Then P ⁢ K ≤ G and exp ⁡ ( G ) is the least common multiple of exp ⁡ ( P ⁢ K ) and exp ⁡ ( H ) , which implies that 𝐺 is not exponent-critical. Hence 𝐻 is minimal non-abelian.

Conversely, suppose that 𝐺 is a direct product of a cyclic 𝑝-group 𝑃 and a minimal non-abelian q ⁢ r -group 𝐻. Any proper subgroup of 𝐺 containing 𝑃 is abelian, since 𝐻 is minimal non-abelian. So no subgroup of 𝐺 is a 𝑝-witness for 𝐺, and hence 𝐺 is exponent-critical. ∎

Let 𝑝 and 𝑞 be distinct primes. Let 𝐺 be a direct product of a normal cyclic Sylow 𝑝-subgroup 𝑃 and a minimal non-abelian 𝑞-subgroup 𝑄. Then 𝐺 is exponent-critical.

Proof

Suppose, for a contradiction, 𝐻 is a 𝑝-witness for 𝐺. Since 𝑃 is cyclic, P ≤ H . Since 𝐻 is proper, H = P × K for some proper subgroup 𝐾 of 𝑄. Since 𝑄 is minimal non-abelian, 𝐾 is abelian. But then 𝐻 is abelian, so 𝐻 is not a 𝑝-witness for 𝐺. This contradiction shows that 𝐺 is exponent-critical. ∎

Let 𝑝 and 𝑞 be distinct primes. Suppose 𝐺 is a non-abelian semi-direct product of a normal abelian Sylow 𝑝-subgroup 𝑃 of 𝐺 by a cyclic 𝑞-complement 𝑄. Suppose that 𝑃 is the direct product of cyclic groups of order p m , where 𝑚 is a positive integer. Furthermore, suppose that | G : C G ( P ) | = q and 𝑄 acts non-trivially and irreducibly on P / P p . Then 𝐺 is exponent-critical.

Proof

Suppose, for a contradiction, that 𝐻 is a 𝑝-witness for 𝐺. We see that H = P 1 ⁢ Q 1 , where P 1 = P ∩ H and where Q 1 is a 𝑞-group. By replacing 𝐻 by a conjugate if necessary, we may assume that Q 1 ≤ Q . Since 𝐻 is non-abelian, 𝑄 is cyclic and | G : C G ( P ) | = q , we see that Q 1 = Q . Since 𝑃 is normal and Q ≤ H , we see that P 1 is normalized by 𝑄. Because 𝐻 has exponent divisible by the 𝑝-part p m of the exponent of 𝐺, the quotient P 1 ⁢ P p / P p is non-trivial. The action of 𝑄 on P / P p is irreducible and P 1 ⁢ P p / P p is 𝑄-invariant, and hence P 1 ⁢ P p / P p = P / P p . So 𝑃 is generated by P 1 and P p , and hence P 1 generates 𝑃 (as P p = Φ ⁢ ( P ) ). Hence P 1 = P . Thus H = P 1 ⁢ Q 1 = P ⁢ Q = G , and so 𝐻 is not proper. This contradiction establishes the lemma. ∎

Let 𝑝 and 𝑞 be distinct primes. Moreover, suppose 𝐺 is a non-abelian semi-direct product of a normal abelian Sylow 𝑝-subgroup 𝑃 of 𝐺 by a cyclic 𝑞-complement 𝑄. For m > 1 , suppose that 𝑃 is the direct product of a cyclic group of order p m and an elementary abelian 𝑝-group. Suppose 𝑄 centralizes the cyclic group of order p m and acts irreducibly on the elementary abelian group. Furthermore, suppose that | G : C G ( P ) | = q . Then 𝐺 is exponent-critical.

Proof

We may write P = C × A , where 𝐶 is a cyclic group of order p m and where 𝐴 is elementary abelian; 𝑄 centralizes 𝐶 and acts irreducibly by conjugation on 𝐴. We choose a generator 𝑦 for 𝑄, so Q = ⟨ y ⟩ .

Suppose, for a contradiction, that 𝐻 is a 𝑝-witness for 𝐺. As in Lemma 3.2, we may assume that H = P 1 ⁢ Q , where P 1 is normalized by 𝑄.

Since 𝐻 is non-abelian, there exists an element g : = c a ∈ P 1 , where c ∈ C and a ∈ A ∖ { 1 } . Now 𝑄 acts non-trivially on 𝐴, since 𝐺 is non-abelian. Since 𝑄 acts irreducibly and non-trivially on 𝐴, we deduce that C A ⁢ ( y ) = { 1 } , and so [ g , y ] ∈ A ∖ { 1 } . Since P 1 is normalized by 𝑄, we see that [ g , y ] ∈ P 1 , and so P 1 ∩ A ≠ { 1 } . But 𝑄 acts irreducibly on 𝐴, and so A ≤ P 1 . Since p m divides the exponent of 𝐻 and since m > 1 , we see that there exists an element x ∈ P 1 such that x ∉ A ⁢ P p . Since A ⁢ P p has index 𝑝 in 𝑃, we find that ⟨ x ⟩ ⁢ A ⁢ P p = P , and so ⟨ x ⟩ ⁢ A = P (since P p = Φ ⁢ ( P ) ). But ⟨ x ⟩ ⁢ A ≤ P 1 , and so P 1 = P . Thus H = P 1 ⁢ Q 1 = P ⁢ Q = G , and so 𝐻 is not proper. This contradiction establishes the lemma. ∎

Let 𝑝 and 𝑞 be distinct primes. Let G = Q ⋊ P be a (non-abelian) semi-direct product of a Sylow 𝑞-subgroup 𝑄 by a non-normal cyclic Sylow 𝑝-subgroup 𝑃. Suppose that 𝑄 is special, so either 𝑄 is elementary abelian or Q ′ = Z ⁢ ( Q ) = Φ ⁢ ( Q ) . Moreover, suppose that 𝑃 acts (by conjugation) trivially on Q ′ and irreducibly on Q / Q ′ . Then 𝐺 is exponent-critical.

Proof

Let G = Q ⋊ P satisfy the conditions of the lemma. We show that there is no 𝑝-witness for exp ⁡ ( G ) .

Assume for a contradiction that 𝐻 is a 𝑝-witness for exp ⁡ ( G ) . Without loss of generality, we may assume that H = Q 1 ⁢ P 1 for some Q 1 ≤ Q and P 1 ≤ P , where Q 1 is P 1 -invariant. Clearly, P 1 contains an element 𝑥 which is of maximal order in 𝑃, and so (since 𝑃 is cyclic) we see that P 1 = P = ⟨ x ⟩ .

Suppose that Q 1 ⊆ Φ ⁢ ( Q ) . Since 𝑄 is special (whether elementary abelian or not), Q ′ = Φ ⁢ ( Q ) ≤ Z ⁢ ( Q ) . So Q 1 ≤ Z ⁢ ( Q ) is abelian and, since 𝑃 centralizes Q ′ , we see that 𝑃 centralizes Q 1 . But then 𝐻 is abelian, and we have a contradiction. We may deduce that Q 1 contains an element in Q ∖ Φ ⁢ ( Q ) , and so Q 1 ⁢ Φ ⁢ ( Q ) / Φ ⁢ ( Q ) is non-trivial.

Since 𝑄 is special, Φ ⁢ ( Q ) = Q ′ , and so 𝑃 acts irreducibly on Q / Φ ⁢ ( Q ) . Since Q 1 is 𝑃-invariant, the non-trivial subgroup Q 1 ⁢ Φ ⁢ ( Q ) / Φ ⁢ ( Q ) is a 𝑃-invariant subgroup of Q / Φ ⁢ ( Q ) . So Q 1 ⁢ Φ ⁢ ( Q ) / Φ ⁢ ( Q ) = Q / Φ ⁢ ( Q ) , and thus Q 1 ⁢ Φ ⁢ ( Q ) = Q . By the non-generation property of the Frattini subgroup, we may deduce that Q 1 = Q , and so it follows that H = G . This contradicts the fact that 𝐻 is proper, as required. ∎

Proof of Theorem C

Lemmas 3.1 to 3.4 show that the four families described in the theorem consist entirely of exponent-critical groups. It suffices to show that any non-abelian exponent-critical p , q -group lies in one of these four families.

Let 𝐺 be a non-abelian exponent-critical group with | G | = p α ⁢ q β , where 𝑝 and 𝑞 are distinct primes. Suppose that there is no 𝑝-witness for 𝐺. So all proper subgroups containing a Sylow 𝑝-subgroup 𝑃 of 𝐺 are abelian, and in particular, 𝑃 is abelian. Moreover, either 𝑃 is normal or N G ⁢ ( P ) is abelian.

Part I: 𝑃 is normal. Suppose that a complement 𝑄 of 𝑃 is also normal, and so G = P × Q . Since no non-abelian proper subgroup of 𝐺 can contain 𝑃, we see that 𝑄 is minimal non-abelian. Now suppose that 𝑃 is not cyclic. Let 𝑥 be an element of maximal order in 𝑃. Then ⟨ x ⟩ ⁢ Q is a 𝑝-witness for 𝐺, which is a contradiction. So 𝑃 must be cyclic and 𝐺 lies in the family described in Lemma 3.1.

Now let us consider the case when 𝑃 is normal but its complement, say 𝑄, is not. So we can find an element y ∈ Q which is not in C G ⁢ ( P ) . Thus P ⁢ ⟨ y ⟩ is a non-abelian subgroup of 𝐺 containing 𝑃. So G = P ⁢ ⟨ y ⟩ and therefore Q = ⟨ y ⟩ . Furthermore, ⟨ y q ⟩ ≤ C G ⁢ ( P ) since otherwise P ⁢ ⟨ y q ⟩ is a 𝑝-witness for 𝐺. So | G : C G ( P ) | = q .

Let the exponent of 𝑃 be p m . Write P ⁢ [ p m − 1 ] for the subgroup of elements of 𝑃 of order dividing p m − 1 , and note that P p ≤ P ⁢ [ p m − 1 ] . Set V = P / P p . We regard 𝑉 as a vector space over F p . Indeed, 𝑉 can be thought of as an F p ⁢ Q -module, with the action of 𝑄 derived from conjugation. Now 𝑄 acts non-trivially on 𝑃 by conjugation, and the only automorphisms of 𝑃 that induce the identity on P / Φ ⁢ ( P ) have 𝑝-power order. Since Φ ⁢ ( P ) = P p , we see that 𝑄 acts non-trivially on 𝑉.

Let π : P → V be the natural homomorphism. Now

is an F p ⁢ Q submodule of 𝑉. Indeed, 𝑈 is a proper submodule, since 𝑃 has exponent p m . Since the order of 𝑄 is coprime to 𝑝, the module 𝑉 is completely decomposable, and so we may write 𝑉 as the sum

of irreducible submodules, where U = U 1 ⊕ U 2 ⊕ ⋯ ⊕ U k and where

forms a complement to 𝑈 in 𝑉. We have k ≥ 0 and, since 𝑈 is proper, ℓ ≥ 1 . Since all the elements in P ∖ P ⁢ [ p m − 1 ] have order p m , we see that π − 1 ⁢ ( W i ) has a subgroup of 𝑃 of exponent p m . We divide our argument into two sub-cases.

Sub-case 1: Suppose that 𝑄 acts non-trivially on a submodule W i . We see that ⟨ π − 1 ⁢ ( W i ) , Q ⟩ is a non-abelian subgroup of 𝐺 of exponent p m , and (since 𝐺 has no 𝑝-witness) we deduce that G = ⟨ π − 1 ⁢ ( W i ) , Q ⟩ . Hence k = 0 , ℓ = 1 and V = W i in this case. Since k = 0 , we see that 𝑃 is a direct product of cyclic groups of order p m . Since W i is irreducible and W i = P / P p , we see that 𝑄 acts irreducibly on P / P p . Thus 𝐺 lies in the family described in Lemma 3.2.

Sub-case 2: 𝑄 acts trivially on all submodules W i . Since 𝑄 acts non-trivially on 𝑉, we see that 𝑄 acts non-trivially on one of the modules U i . (In particular, this implies that 𝑄 acts non-trivially on 𝑈, and so m > 1 .) The subgroup ⟨ π − 1 ⁢ ( U i ⊕ W 1 ) , Q ⟩ is non-abelian and of exponent p m , and so

Hence k = 1 = ℓ and V = U 1 ⊕ W 1 . If 𝑄 acts non-trivially by conjugation on P p , the subgroup ⟨ π − 1 ⁢ ( W 1 ) , Q ⟩ would be a 𝑝-witness, and so we deduce that 𝑄 centralizes P p .

Taking 𝑝-powers in 𝑃 induces a surjective F p ⁢ Q -module homomorphism 𝑓 from V = P / P p to the F q ⁢ Q -module P p / P p 2 . Since m > 1 , this homomorphism is non-trivial. Since 𝑄 centralizes P p / P p 2 and acts non-trivially and irreducibly on U 1 , we see that f ⁢ ( U 1 ) = 0 and f ⁢ ( W 1 ) = P p / P p 2 . Since W 1 is trivial and irreducible, it has dimension 1, and so 𝑃 is a direct product of a cyclic group of order p m with an elementary abelian group.

Let x ∈ π − 1 ⁢ ( W 1 ∖ { 0 } ) . The element x ∈ P has order p m and (since | Q | is coprime to 𝑝) 𝑥 is centralized by 𝑄. The action of 𝑄 by conjugation on P ⁢ [ p ] gives P ⁢ [ p ] the structure of an F p ⁢ Q -module. Let 𝐴 be a complement to the submodule in P ⁢ [ p ] generated by x p m − 1 . Then P = ⟨ x ⟩ × A , where the action of 𝑄 fixes 𝑥 and preserves the direct product. Since A ⁢ P p / P p = U 1 , we see that the action of 𝑄 on 𝐴 is irreducible. So 𝐺 lies in the family described in Lemma 3.3.

Part II: 𝑃 is not normal, and so N G ⁢ ( P ) is abelian. Clearly, in this case, we have P ≤ Z ⁢ ( N G ⁢ ( P ) ) and by Burnside’s Normal 𝑝-Complement Theorem, 𝐺 has a normal 𝑝-complement 𝑄. Therefore, 𝐺 is a semi-direct product of its Sylow 𝑞-subgroup 𝑄 by a non-normal abelian Sylow 𝑝-subgroup 𝑃.

Suppose, for a contradiction, that 𝑃 is not cyclic. Let x ∈ P have maximal order. Then ⟨ x ⟩ ⁢ Q is a proper subgroup, so must be abelian (otherwise, we have a witness for the 𝑝-part of exp ⁡ G ). So all elements of maximal order lie in C G ⁢ ( Q ) . But 𝑃 is generated by its elements of maximal order, and so P ≤ C G ⁢ ( Q ) . This implies that 𝑃 is normal, and we have our contradiction. Hence 𝑃 is cyclic.

The subgroup 𝑃 must centralize any proper 𝑃-invariant subgroup of 𝑄; otherwise, we have a witness to the 𝑝-part of exp ⁡ ( G ) . So a generator 𝑥 of 𝑃 gives rise (via conjugation) to an automorphism of 𝑄 that acts trivially on any proper subgroup of 𝑄. This automorphism is non-trivial, since 𝑃 is not normal. Hence, by [8, Theorem 5.3.7], 𝑄 is special, 𝑃 acts trivially on Q ′ , and 𝑃 acts irreducibly on Q / Q ′ . So 𝐺 lies in the family described in Lemma 3.4, and the theorem follows. ∎

Let 𝑃 be a finite non-abelian exponent-critical 𝑝-group. Then we have P = ⟨ a , b ⟩ , where a ∈ P has maximal order and b ∈ P . Moreover, 𝑃 is solvable of derived length 2.

Proof

The Frattini subgroup Φ ⁢ ( P ) is contained in all maximal subgroups. Since at least one maximal subgroup is abelian, it follows that Φ ⁢ ( P ) is abelian. Since P ′ ≤ P ′ ⁢ P p = Φ ⁢ ( P ) and since 𝑃 is non-abelian, 𝑃 is solvable of derived length 2.

We now show that 𝑃 is 2-generated. Suppose, for a contradiction, that all elements in M P are central. Let a ∈ M P , and let x , y ∈ P be such that [ x , y ] ≠ 1 . As 𝑥 is not central, it does not have maximal order. Hence, using the fact that [ a , x ] = 1 , we see that x ⁢ a has maximal order. Since 𝑎 is central,

Thus we have a contradiction as required, and we may deduce that there exists an element a ∈ M P of maximal order that is not central. Let b ∈ P be such that [ a , b ] ≠ 1 . Thus ⟨ a , b ⟩ is a non-abelian subgroup of exponent exp ⁡ ( P ) , and so P = ⟨ a , b ⟩ . Hence the theorem follows. ∎

Proof of Theorem D

Suppose 𝑃 is a non-abelian finite group which is 2-generated and | P ′ | = p . We show that 𝑃 is exponent-critical of type ℬ. Let P = ⟨ a , b ⟩ . Since 𝑃 is nilpotent, [ P ′ , P ] is a proper subgroup of P ′ . Thus 𝑃 has nilpotency class 2. In particular, P ′ ≤ Z ⁢ ( P ) . Since 𝑃 has nilpotency class 2, we get that 1 = [ a , b ] p = [ a p , b ] = [ a , b p ] . So a p and b p are central. Hence Φ ⁢ ( P ) ≤ Z ⁢ ( P ) as, in a 𝑝-group, Φ ⁢ ( P ) is the subgroup generated by P ′ and the 𝑝-th powers of a generating set. Let 𝑀 be a maximal subgroup. Then Φ ⁢ ( P ) ≤ M . Since 𝑃 is 2-generated, Φ ⁢ ( P ) has index p 2 in 𝑃 and so has index 𝑝 in 𝑀. Since Φ ⁢ ( P ) is central in 𝑃, we see that Z ⁢ ( M ) ≥ Φ ⁢ ( P ) , and so the center of 𝑀 has index at most 𝑝 in 𝑀. But then 𝑀 must be abelian (and Z ⁢ ( M ) = M ). Thus 𝑃 is a minimal non-abelian group and hence exponent-critical of type ℬ, as required.

Conversely, suppose that 𝑃 is exponent-critical of type ℬ. By Theorem 4.1, P = ⟨ a , b ⟩ for elements a , b ∈ P with a ∈ M P . Since 𝑃 has type ℬ, it contains distinct abelian maximal subgroups M 1 and M 2 . For x , y ∈ P , we see that [ x , y ] ∈ P ′ ≤ Φ ⁢ ( P ) ≤ M 1 ∩ M 2 , and so [ x , y ] commutes with all elements of M 1 and M 2 . Since P = M 1 ⁢ M 2 , we deduce that P ′ is central in 𝑃, and so 𝑃 has nilpotency class 2. Hence [ x , y ⁢ z ] = [ x , y ] ⁢ [ x , z ] and [ x ⁢ y , z ] = [ x , z ] ⁢ [ y , z ] for all x , y , z in 𝑃. Using this, we see that P ′ = ⟨ [ a , b ] ⟩ . Let M a be a maximal subgroup of 𝑃 such that a ∈ M a . Since 𝑎 has maximal order and 𝑃 is exponent-critical, we see that M a is abelian. Clearly, b p ∈ M a . Since 𝑃 has nilpotency class 2, we see that 1 = [ a , b p ] = [ a , b ] p = [ a p , b ] . So o ⁢ ( [ a , b ] ) = p = | P ′ | , and part (iii) of the theorem follows. ∎

Let 𝑃 be an exponent-critical finite non-abelian 𝑝-group of type ℬ of order p n . Then

where α , β , ρ , σ are integers such that α ≥ β ≥ 1 , α + β = n − 1 , 0 ≤ ρ , σ ≤ 1 . When 𝑝 is odd, 𝑃 is isomorphic to exactly one of the groups whose parameters ( α , β , ρ , δ ) are listed below:

1. 1. ( α , β , 0 , 1 ) with α > β ≥ 1 ,

   2. ( α , β , 1 , 1 ) with α > β ≥ 1 ,

   3. ( α , β , 1 , 0 ) with α > β ≥ 1 ,

2. 1. ( α , α , 0 , 1 ) with α ≥ 1 ,

   2. ( α , α , 1 , 1 ) with α ≥ 1 .

1. 1. ( α , β , 0 , 1 ) with α > β ≥ 1 ,

   2. ( α , β , 1 , 1 ) with α > β ≥ 1 ,

   3. ( α , β , 1 , 0 ) with α > β ≥ 1 ,

2. 1. ( α , α , 0 , 1 ) with α > 1 ,

   2. ( α , α , 1 , 1 ) with α > 1 ,

3. 1. ( 1 , 1 , 0 , 0 ) ,

   2. ( 1 , 1 , 1 , 1 ) .

Let 𝑃 be a non-abelian exponent-critical 𝑝-group of type 𝒜. Let 𝐴 be a maximal subgroup containing an element of maximal order p m .

1. 𝐴 is abelian, normal, and contains all elements of order p m .

2. A ≅ Z p m × S , where 𝑆 has exponent dividing p m − 1 .

3. P ′ is abelian, of exponent dividing p m − 1 .

Proof

𝐴 is normal, since all maximal subgroups of a 𝑝-group are normal. If 𝐴 were non-abelian, 𝑃 would have a proper non-abelian subgroup of exponent p m and hence 𝑃 would not be exponent-critical. This contradiction shows that 𝐴 is abelian. Since 𝑃 is of type 𝒜, we have that 𝐴 is the unique maximal subgroup of 𝑃 which is abelian. Clearly, 𝐴 will contain all the elements of maximal order and part (i) of the lemma follows.

To prove part (ii), first note that 𝐴 has exponent p m , and so A ≅ ( Z p m ) r × S , where 𝑟 is a positive integer and 𝑆 has exponent dividing p m − 1 . Define 𝐾 to be the subgroup of all elements of 𝐴 of order dividing p m − 1 . Every element of A ∖ K has order p m , and A / K is elementary abelian of order p r . Let b ∈ P ∖ A . Since 𝑏 acts nilpotently on A / K , there exists a subgroup 𝐿 containing 𝐾 such that b ∈ N P ⁢ ( L ) and 𝐿 has index 𝑝 in 𝐴. Note that 𝐿 is normal in 𝑃, since { b } ∪ A ⊆ N P ⁢ ( L ) . Consider the subgroup 𝑀 generated by 𝐿 and 𝑏. Since 𝐴 has index 𝑝 in 𝑃, we get that b p ∈ A . Since b ∈ P ∖ A , the order of 𝑏 divides p m − 1 , and so b p ∈ K ≤ L . Hence 𝑀 has index 𝑝 in 𝑃 and so is maximal. If r > 1 , then 𝐿 contains elements of A ∖ K , and so 𝐿 contains an element of maximal order. Hence 𝑀 must be abelian (as 𝑃 is exponent-critical) and therefore M = A . But b ∈ M and b ∉ A . This contradiction shows that r = 1 , and so part (ii) is established.

To prove (iii), we note that 𝐾 has index 𝑝 in 𝐴 by part (ii) and hence 𝐾 has index p 2 in 𝑃. So P / K is abelian and therefore P ′ ≤ K . Hence the exponent of P ′ divides p m − 1 , as required. ∎

Let 𝑃 be a 𝑝-group. Suppose 𝑃 possesses an abelian maximal subgroup 𝐴. Let b ∈ P ∖ A .

1. The map κ : A → A defined by κ ⁢ ( x ) = [ x , b ] is a homomorphism.

2. For x ∈ A , and b ∈ P ∖ A ,

   (4.1) ( b x ) i = b i x i ∏ j = 1 i − 1 [ x , j b ] ( i j + 1 ) .

3. For x ∈ A , and b ∈ P ∖ A ,

   [ x , p b ] = ∏ i = 1 p − 1 [ x , i b ] − ( p i ) .

Proof

To prove (i), we note that, for all x ∈ A , [ x , b ] ∈ A as 𝐴 is normal in 𝑃. So [ x , b ] y = [ x , b ] for all y ∈ A , since 𝐴 is abelian. Hence

and 𝜅 is a homomorphism, as required.

We prove equality (4.1) by induction on 𝑖. Equality (4.1) clearly holds when i = 1 . Assume the equality holds for a fixed value of 𝑖. Since P / A is abelian, all commutators lie in the abelian subgroup 𝐴. Hence

and so part (ii) follows by induction.

Finally, we prove part (iii) of the lemma. We first prove that, for any positive integer 𝑖,

To show this, we note that b − 1 ⁢ x ⁢ b = x ⁢ [ x , b ] , and so (4.2) holds when i = 1 . If (4.2) holds for a given value of 𝑖, we see that

and so, by induction, equation (4.2) holds for all 𝑖. Now b p ∈ A since 𝐴 is maximal. As 𝐴 is abelian, b − p ⁢ x ⁢ b p = x , and so

Hence part (iii) of the lemma follows. ∎

The automorphism 𝜙 defined in Definition 1.3 has order 𝑝. In particular, the definition of 𝑈 as a semi-direct product is well-defined.

Proof

We see that, for 0 ≤ j ≤ p − 1 ,

and so a short calculation shows that ϕ p ⁢ ( a 0 ) = ϕ ⁢ ( ϕ p − 1 ⁢ ( a 0 ) ) = a 0 . The lemma now follows since 𝑊 is generated by ϕ j ⁢ ( a 0 ) for 0 ≤ j ≤ p − 1 . ∎

Let U = W ⋊ ⟨ b 0 ⟩ be the semi-direct product defined above.

1. We have a i = [ a 0 , i b 0 ] for 0 ≤ i ≤ p − 1 . In particular, U = ⟨ a 0 , b 0 ⟩ .

2. We see that | U | = p ( m − 1 ) ⁢ ( p + 1 ) + 1 .

3. The group 𝑈 is defined by the relations

   [ [ a 0 , i b 0 ] , [ a 0 , j b 0 ] ] = 1 for ⁢ 0 ≤ i < j ≤ p − 1 , [ a 0 , p , b 0 ] = ∏ j = 1 p − 1 [ a 0 , j b 0 ] − ( p j ) , a 0 p m = 1 , [ a 0 , i b 0 ] p m − 1 = 1 for ⁢ 1 ≤ i ≤ p − 1 , b 0 p m − 1 = 1

   in the generating set { a 0 , b 0 } .

4. The group 𝑈 is solvable of class 2. The subgroup ⟨ b 0 p ⟩ ⁢ W is abelian and maximal.

5. The derived subgroup U ′ of 𝑈 is given by

   U ′ = ⟨ a 1 , a 2 , … , a p − 1 ⟩ = ⟨ [ a 0 , i b 0 ] : 1 ≤ i ≤ p − 1 ⟩ .

   In particular, U ′ is abelian of exponent p m − 1 and order p ( m − 1 ) ⁢ ( p − 1 ) .

6. The Frattini subgroup Φ ⁢ ( U ) of 𝑈 is given by

   Φ ⁢ ( U ) = ⟨ a 0 p , a 1 , a 2 , … , a p − 1 , b 0 p ⟩ .

7. Any non-trivial subgroup of U ′ that is normal in 𝑈 must contain the (non-trivial) element [ a 0 , p − 1 b 0 ] p m − 2 .

Proof

Parts (i) to (vi) of the lemma are straightforward to prove, and we leave them to the reader. Note that Lemma 4.6 (iv) allows us to use Lemma 4.4 when reasoning about 𝑈.

To prove (vii), note that the map κ : U ′ → U ′ defined by κ ⁢ ( x ) = [ x , b 0 ] is a homomorphism with kernel ⟨ a p − 1 p m − 2 ⟩ of order 𝑝. Since 𝑈 is nilpotent, κ i is trivial for some positive integer 𝑖. Let K ≤ U ′ be a non-trivial subgroup of U ′ that is normal in 𝑈. Let 𝑖 be the smallest integer such that κ i ⁢ ( K ) = { 1 } . Then

Since 𝐾 is normal in 𝑈, we have κ i − 1 ⁢ ( K ) ≤ K , and so part (vii) of the lemma follows. ∎

Proof of Theorem E

Let 𝑃 be a non-abelian exponent-critical 𝑝-group of exponent p m of type 𝒜. We prove part (i) of the theorem by showing that P ≅ U / N for some normal subgroup N ∈ N .

By Theorem 4.1, we have that P = ⟨ a , b ⟩ , where a , b ∈ P with 𝑎 of order p m . Consider the free group 𝐹 generated by a 0 and b 0 . Let 𝜋 be the (surjective) homomorphism from 𝐹 to 𝑃 mapping a 0 to 𝑎 and b 0 to 𝑏. We show that 𝑃 is a quotient U / N of 𝑈 by checking that the relations defining 𝑈 are all satisfied under 𝜋. (Here we are making use of the fact, see Lemma 4.6 (i), that U = ⟨ a 0 , b 0 ⟩ .)

Let 𝐴 be the abelian maximal subgroup containing 𝑎. Since 𝑃 is not abelian, b ∈ P ∖ A . Lemma 4.3 (i) shows that 𝑏 has order dividing p m − 1 , since all elements of order p m lie in 𝐴. Hence a p m = b p m − 1 = 1 . Since P ′ ≤ Φ ⁢ ( P ) ≤ A , we see that the subgroup generated by P ′ and 𝑎 is abelian. In particular,

Lemma 4.3 (iii) implies that [ a 0 , i , b 0 ] p m − 1 = 1 for 1 ≤ i ≤ p − 1 . Finally, Lemma 4.4 (iii) shows that [ a 0 , p b 0 ] = ∏ j = 1 p − 1 [ a 0 , j b 0 ] − ( p j ) . Hence P ≅ U / N for some normal subgroup 𝑁. Moreover, under this isomorphism, the element a ∈ P corresponds to the coset a 0 ⁢ N , and 𝑏 corresponds to b 0 ⁢ N .

We now check that the normal subgroup 𝑁 lies in 𝒩. Since a 0 and 𝑎 both have order p m , we see that N ∩ ⟨ a 0 ⟩ = { 1 } . Since 𝑃 is non-abelian, we see that U ′ ≰ N . Since 𝑈 and 𝑃 are 2-generated, we see that N ≤ Φ ⁢ ( U ) . So to prove part (i) of the theorem, it remains to show that D ≤ N . It suffices to show that a p m − 1 [ a , p − 1 b ] p m − 2 = 1 in 𝑃.

Consider the element b ⁢ a ∈ P ; since b ⁢ a ∉ A , we see that b ⁢ a has order dividing p m − 1 , by Lemma 4.3 (i). Now, by Lemma 4.4 (ii),

since 𝑝 divides all the binomial coefficients in the product and since all the factors lie in the abelian subgroup 𝐴. But 𝑏 has order dividing p m − 1 and the commutators all have order dividing p m − 1 by Lemma 4.3 (iii). Hence ( b a ) p = a p [ a , p − 1 b ] x , where x ∈ A has order dividing p m − 2 . So

as required. Hence part (i) follows.

To prove part (ii), we first investigate the orders of elements in the quotient Q = U / D . Since D ∩ ⟨ a 0 ⟩ = { 1 } , the element a 0 ⁢ D ∈ Q has order p m . Hence the exponent of 𝑄 is divisible by p m . Define 𝑀 to be the subgroup M : = ⟨ b 0 p ⟩ W . By Lemma 4.6 (iv), 𝑀 is maximal and abelian of exponent p m . Since D ≤ M , the subgroup M / D is maximal in 𝑄. Moreover, since a 0 ⁢ D ∈ M / D , we see that M / D of 𝑄 has exponent p m . Our next aim is to show that

To see this, write g ∈ Q ∖ ( M / D ) in the form g = b 0 r ⁢ x ⁢ D , where 1 ≤ r ≤ p − 1 and x ∈ M . Replacing 𝑔 by g r − 1 mod p does not change the order of 𝑔, and so we may assume without loss of generality that r = 1 . Since M / D is an abelian maximal subgroup of 𝑄, Lemma 4.4 (ii) shows that

The factors in this product all lie in the abelian subgroup M / D of 𝑄. The factor b 0 p ⁢ D has order dividing p m − 2 as b 0 has order p m − 1 . The factors [ x , j b 0 ] ( p j + 1 ) D for 1 ≤ p − 2 have order dividing p m − 2 , since (by Lemma 4.6 (v)) the derived subgroup of 𝑈 has exponent p m − 1 and ( p j + 1 ) is divisible by 𝑝. Hence

Define

The map x ↦ x p m − 1 is a homomorphism on 𝑀. Moreover, the map x ↦ [ x , b 0 ] is a homomorphism on 𝑀 by Lemma 4.4 (i), and so the map x ↦ [ x , p − 1 b 0 ] is also a homomorphism. Thus the set 𝑆 is in fact a subgroup of 𝑀. To prove (4.3), it suffices to show that S = M . Let x = [ a 0 , i b 0 ] , where 1 ≤ i ≤ p − 1 . Then 𝑥 has order dividing p m − 1 since x ∈ U ′ . Now, the order of [ x , p − 1 b 0 ] divides the order of [ x , p − i b 0 ] . But the order of [ x , p − i b 0 ] divides p m − 2 since