Let G be a finite 2-group with the property that for all subgroups H of G. Then G has an abelian normal subgroup of index at most 4 in G. This result represents an affirmative answer to Question 18.56 from the current edition of the Kourovka Notebook.
Let n be a positive integer. A group G is called core-n if for every subgroup H of G; here, denotes the largest normal subgroup of G contained in H. In , it was shown that a locally finite core-n group has a normal abelian subgroup whose index in the group is bounded by a function of n. In  the same authors proved that, for odd primes p, a finite core-pp-group has an abelian normal subgroup of index , a best possible bound. The paper ([3, Theorem 2]) also contains a proof of the existence of a normal abelian subgroup of index at most in a finite core-2 2-group. The authors mention that no examples of finite core-2 2-groups were known that did not possess an abelian subgroup of index at most 4. Later () it was shown that finite core-2 groups of class 2 indeed must have an abelian subgroup of index at most 4. The present paper is devoted to showing that every finite core-2 2-group has an abelian subgroup of index at most 4. This provides an affirmative answer to Question 18.56 from the 18th edition of the Kourovka Notebook, posed by G. Cutolo.
The dihedral groups are core-2, which shows that core-2-ness does not entail bounded class or bounded subgroup breadth. This places core-2 2-groups in contrast to odd-order core-pp-groups, those being of class at most 3 (). The group is core-2, which shows that 4 is the best possible upper bound for the index of an abelian subgroup in a core-2 2-group. Our main result reads as follows:
Let G be a finite 2-group. Suppose that for every subgroup H of G. Then G has an abelian subgroup that contains and has index at most four in G.
We note that the corresponding result for odd p is an immediate corollary of the results of . Letting p be an odd prime and G a finite core-pp-group, G has a normal abelian subgroup A of index ([3, Theorem 1]). We may obviously take A to be a maximal abelian subgroup of G. Then, according to [3, Lemma 1.1 (iii)], and , because is too small to be nonabelian.
We conclude this introduction with three very helpful observations: Subgroups as well as quotients of a core-n group are core-n. If G is a core-pp-group, then whenever and if E is an elementary abelian subgroup of G. We also note a corollary.
Let G be a finite core-2 group. Then G has an abelian subgroup of index at most 8, and this bound is best possible.
Let G be a finite core-2 group of composite order, and let . Then Q is abelian of odd order, every subgroup of Q is normal in G and with . The group S is core-2, and, since , has index at most 2 in S. A direct product of a core-2 2-group with an abelian group of odd order is core-2; if and , then the corresponding semidirect product SQ is core-2 if and only if S normalises every subgroup of Q, , S is core-2 and whenever . By the theorem, T has an abelian subgroup of index at most 4 and the direct product of this with Q has index at most 8 in G. Now let
Then is core-2, but there is no abelian subgroup of index 4 in S that does not contain y. Now let , where , and y inverts the elements of Q. This G does not have an abelian subgroup of index four. ∎
The reader interested in the structure of core-p-groups of class 2 may find the results of Lemma 9 somewhat relevant.
Lemma 1 ([2, Lemma 3.4.1])
Let X be a finite group and let B be an abelian normal subgroup of X. Let N be the subgroup of X consisting of elements inducing power automorphisms on B. Let and let U be the unit subgroup of . There is a homomorphism given by whenever and .
Let X be a finite group. Following Blackburn’s notation, we let
Lemma 2 ([1, Satz 1])
Let P be a 2-group and let . Let be the set of maximal subgroups of N. Then
The fact that core-pp-groups have abelian Frattini subgroup is proved in [3, see Lemma 1.1 (ii)–(iii) for odd p and Proposition 2.11 for ]. We include a short proof for , as this will be a vital ingredient of the forthcoming proof.
Lemma 3 ([3, Proposition 2.11])
A finite core-2 2-group P has abelian Frattini subgroup.
Let P be a counterexample of minimal order. Since whenever , is cyclic and . Let s and t be involutions in . If , then . Now P normalises a maximal subgroup V of , of order 4; this, however, implies that , a contradiction. Let . We have just seen that Y is elementary abelian, whence , implying and . Let . If , then . Since for every x in P of order greater than 2, a subgroup U of not containing z must be contained in Y, hence in . Thus is Hamiltonian, i.e. with E elementary abelian and . At least one maximal subgroup V of Q must be normal in P, such that , a contradiction. ∎
Let P be a finite p-group. Let U and V be subgroups of P satisfying . If for all v in V, then or .
Let and . If and , then , of order . It follows that every element of V belongs to or belongs to the subgroup of V. The assertion follows because V cannot be the union of two proper subgroups. ∎
A proof of the following lemma could be compiled from references to the available literature on metacyclic groups.
Let be a 2-group with . Unless , there is with .
Assume that ; it will suffice to establish the corresponding statement in the quotient . Hence assume with and . Observe that this implies . Let . A generator w of may be chosen such that is equal to a power ; as and , we have .
Assuming , let . Note that the order of in is equal to . Since and ,
It follows that and . However,
We have established that . Letting , we find that
So unless . Now implies . Hence if , then . Accordingly, with and . However, yields . ∎
As usual, “” denotes “maximal subgroup of” and the minimal number of generators of the p-group P.
Observe that the core-p-ness of a p-group G is already guaranteed if is required for all subgroups of G generated by (no more than) two elements. Indeed, suppose for all elements of the p-group G. This property carries over to subgroups and quotients. Let and let . Subgroups of are core-free, and if , then has a 2-generated subgroup of order greater than p, a contradiction. Another equivalent characterisation of core-p-ness may be found in [3, Lemma 1.3]: The group G enjoys core-p-ness if and only if every nontrivial subgroup H of G has a maximal subgroup M with .
For the remainder of this paper, p is a prime and G denotes a finite core-pp-group of order greater than p. The equivalent formulation of core-p-ness of a p-group predominantly used in the proofs to come is a mixture of the two characterisations given in the previous paragraph: Every nontrivial subgroup of a core-pp-group generated by two elements has a maximal subgroup that is normal in the group.
Observe that whenever and that for each elementary abelian subgroup E of G.
Let . For and , the involution in is denoted by .
Although our theorem only concerns the prime 2, a few lemmas in this section will be stated for all primes. This was done where it would not increase the length of the respective proof inordinately and it was felt that the information contained in the lemma could be of some interest beyond the requirements of the present paper.
Let A be an abelian normal subgroup of G such that and not every subgroup of A is normal in G. Let be the minimal order of a cyclic subgroup of that is not a normal subgroup of . Then .
Let a be any element of A of order and let . Let be a complement of in . Note that . There is satisfying . Both and are normal in G, while . Thus , i.e. . ∎
Assume . Then unless , and , where E is elementary abelian, , , and with , and .
The commutator map induces a symplectic form on , whence is elementary abelian of even degree. Let .
Assume that p is odd. If , then G is a direct product with Q extraspecial and E elementary abelian. Letting , m is the degree of a maximal elementary abelian core-free subgroup of Q, i.e. .
Now assume . If is noncyclic, then there is such that and the theorem follows by induction. Hence may be supposed to be cyclic. There is of order . Since , , so . Since p is odd, raising elements to their pth power is an endomorphism of G, such that and . Hence there is with . Let , noting that . For , is core-free elementary abelian of order . Thus whenever . Yet if satisfies , then and with , i.e. . Since is cyclic, Q is abelian after all.
Now let , let and let have order .
We first prove the lemma making the additional assumption that is cyclic. Then is a cyclic group generated by elements of order at most , i.e. . Let and , observing that and if and only if Q is abelian.
Assume that . For each , there is a generator w of such that with some . Either , or , and . In particular, y may be chosen of order 2. Assume that Q is nonabelian. If , then an analogous argument yields an involution s in . However, this would mean . Thus and, for , we have and there is an involution satisfying . If , then and .
We turn to the case . Then with D extraspecial and E abelian. The group is not core-2, whence either or and E is elementary abelian, a scenario covered by the assertion.
From now on, is assumed to be noncyclic; observe this implies that . Let and . Then , and is immediate by induction on the group order. Induction also yields that if , then whenever . Thus implies .
Now assume that , and . Let
Then , in particular . Let with , let and let .
For and , is an abelian subgroup of G with . Suppose that . If , then . Consequently, we have with E elementary abelian, , and . Now let . Then and , so .
We have seen that whenever . We shall show that . Assume otherwise. Let . If , then , so . Since , it follows that t may be assumed to satisfy . Let and . Then , while would imply . Hence we have , i.e. . Now let . Then , a contradiction.
We have confirmed the existence of an involution v in the set . Since , the element u may be assumed to satisfy . Let and let . For , , so with and elementary abelian E. This completes the proof. ∎
The elementary abelian group of order will, as usual, be denoted by .
Let and . Following , we define , the breadth of x in U, as . We write instead of . The breadth in G of the subgroup A is defined as .
Let . For a subgroup U of N, let be the preimage of in G.
Suppose that and . Then for . For any subgroup Wof N such that and , there is at most one maximal subgroup M of N with and .
The first assertion is a straightforward application of Lemma 7 applied to , where .
For the second assertion, let and let with and . Note that W is contained in exactly three maximal subgroups of N.
Let M be a maximal subgroup of N with . Let and let . By Lemma 7, and . Assume that . By Lemma 7, , where , and with and . Letting and , we have . Accordingly, . Certainly for any maximal subgroup M of N not containing . So if for at least two maximal subgroups of N containing W, then
which Blackburn’s formula (Lemma 2) makes impossible. Thus .
We note: For , , , , and .
Let . From Lemma 7 and , we deduce that if and only if . From it also follows that, if satisfies , then is isomorphic to one of the groups , , or . Suppose that for at least two maximal subgroups M containing W. Then, as we have seen,
However, , and Blackburn’s formula is good for another contradiction. ∎
The assertions of the next lemma, describing finite core-pp-groups G with , are partly contained in [3, Theorem 1] and [5, Theorem]. We shall, however, require more detail than is provided in the referenced sources.
Suppose that . Let A be an abelian subgroup of G of maximal order. Then the following hold:
If p is odd, then and if , then .
If , then either or . If , then with , in particular .
If and , then or .
Note that .
We start with the case p odd. Let of order , and let of order . For any maximal subgroup U of W, there is a maximal subgroup of N satisfying . Applying Lemma 7 to the quotient , we obtain . Note that this implies that .
Let be the set of pairs , where , and with and . Let . We list the subgroups of V of order p as . For , let and let . We assume the listed as to satisfy for all i. We count the elements of in two ways, to obtain
Suppose that . Then
However, , whence
We have established that either or there is such that contains every element of A of breadth greater than 1. Assume the latter together with . Let . Then
and , while . Let and . Then and , i.e. , a contradiction. Thus or .
Now let . Suppose that A has a subgroup such that and . First suppose that there are subgroups of and of such that . Let and let V be a complement of U in N. Let , , and . For , let . For each index i, we have , while and . Let . Let ; then neither a nor b belongs to . Now Lemma 8 may be invoked to yield that for at least two out of three different choices of X; thus
The intersection over all complements of U in N being trivial, we obtain that in contrast to our assumptions on . If has a subgroup U of order 4, then, similarly, whenever , which results in . The only remaining possibility is that , while ; let and , , , , . Lemma 8 again yields
and a contradiction ensues as before.
Suppose that A has an element a of breadth greater than 1. As seen in the two preceding paragraphs, this implies that , where and for every b in B that satisfies . In particular, .
Assume that . Applying Lemma 4, we obtain or . For b in , we have , which implies and .
Suppose that . If , then there are b and c in B with . Note that
For , we have because of and . Furthermore, is a complement of in . Since there are only two such complements, b and c may be chosen such that . However, this yields , a contradiction. Consequently, entails that with , while , i.e. .
We turn to the case . Let . We have and . Letting , we find that
Given with , we obtain that , a contradiction. Accordingly, and if , then once again , i.e. .
Letting p be any prime, assume . Let , i.e. . It will suffice to derive a contradiction from . For , the maximality of implies that is a maximal abelian subgroup of G. Let and let . One of the maximal subgroups of is normal in G, say . Since , i.e. , we must have . If , then , so and . It follows that there is with . This implies that and there is with , a contradiction.
The final case left for us to consider is . Then Lemma 4 yields or . Only the first case needs further consideration. Let , of order p. We note that for we have or . Let , where z projects nontrivially into along the given decomposition. Let . Then ; however, this implies that and or with , i.e. .
At this point, only the final assertion of the lemma has not been taken care of. By way of contradiction, assume that , and . Recall that .
Let D be a subgroup of of maximal order subject to possessing a complement in A and let . Setting , either , or some c of the form with belongs to . For such a c, has a complement in E, whence has a complement in G. It follows that and with . Note that .
The following observation is going to be quite helpful: Suppose there is such that . Since , induction yields that or there is a maximal subgroup B of G with . However, , whence and .
Let . Then .
Suppose that . Let . Any satisfying then belongs to , and () yields .
Assume that . Since there is with , this implies that . As we just saw, this means , in particular . Let . There is satisfying , i.e. . Since , and the coset wA contains an involution. Letting , it follows that with involutions x and y satisfying , , , and . Now , for none of the elements is in . Accordingly, we have , and, since a maximal subgroup of is normal in G, we must have . This implies and Lemma 7 delivers a contradiction. So .
Assume that . For , let be the preimage of in A. Since , Lemma 7 yields for each v. Since , whenever , and it follows that each of the three maximal subgroups of A that contain is equal to some , i.e. . Now Lemma 4 yields , a contradiction.
We have found that . In particular, , which we have seen to imply . Write with . If , then . As previously mentioned, there is an element with . Since , it follows that , and () is contradicted. Thus , and since , we have .
Suppose that . There is an element satisfying . Since , condition () yields a contradiction. Hence and .
Let . We have seen that , and Lemma 7 says that the subgroup is of index at most 2 in . Only two of the three elements have breadth 1 in G, so there is an element with . Letting and , we obtain a final contradiction. ∎
The following three auxiliary lemmas each dispense with some scenario that makes repeated appearances during coming proofs.
Let P be a finite 2-group having a maximal abelian normal subgroup . If
, , ,
, , ,
then P is not core-2.
Assume otherwise. The four conditions (a)–(d) entail and with , , , , and . Without loss, . Neither of the subgroups and being normal in P, so is, consequently,
For , therefore satisfies (a)–(d) and induction yields a contradiction. Accordingly,
Since E is a complement of in B and contained in , condition (2.2) yields . Note that , for otherwise. Certainly . Set .
Assume that . It follows that and there is such that . Replace u by . If both i and are odd, then and , i.e. . If i is odd and is not, then has an element whose square is a, making normal in P, which it is not. Accordingly, we may take . Since , none of the maximal subgroups of is normal in P. This proves
If i is odd, then (2.1) yields that , i.e. must be even; yet that implies there is with , i.e. . Hence . If or j is even, then there is satisfying , and none of the maximal subgroups of is normal in P.
The only possibility left standing is and . If is odd, then, as before, , so, replacing u by an appropriate element of , we obtain . Since this means , one of and is normal and there is no loss in taking . By (2.3), the situation under consideration is symmetric in b and d, so we may also assume and . Let . Then and , a final contradiction. ∎
Let P be a finite 2-group with a normal subgroup Q such that the following conditions are satisfied:
, where and , while .
with for .
Then P is not core-2.
It will be helpful to be aware of the fact that whenever and . Let . Combining conditions (a) and (b), we see that , while whenever ; in particular for every . The condition placed on the orders of the in (b) entails that . Let . Then will be called an admissible triple if, for , has each of the properties ascribed to in conditions (a) and (b). Note that this implies for . Next, let . By (c), and whenever , so x inverts every element of the abelian group . It follows that . Thus condition (a) entails that or .
Suppose that P is a core-2-group. First assume . Let and let satisfy . If , then , and (d) yields , which is not compatible with (a) and (b). Thus and there is an involution s in . By (d), , whence . For with , we have ; either possibility yields , which contradicts . Consequently:
, in particular .
For , let , in other words . By condition (c), , while . Let . Then it follows from (b) and (c) that is abelian of degree at least 2 and neither of the groups and can be normal in P. Thus:
If , then .
We intend to show that . Since x is an arbitrary element of , that will, Q being nonabelian, suffice to establish a contradiction. So let us assume . Let satisfy ; by condition (2)*, . Let . Since , we have with and . For , , and condition (2)* implies that , while . Let . Since and , condition (2)* yields:
Let . Unless , .
Every element of Q of order less than belongs to , so we know . Suppose that . Then , where and is an admissible triple. Suppose that . By condition (3)* this means that whenever . Accordingly, , since otherwise some w in would satisfy . So or . If , then there is an element with , while of course is admissible. If , then is admissible, while, by (3)* and because of , if and only if if and only if if and only if .
We have found an admissible triple with and . Since and is admissible, we may even take . Recall that . From we get . Since , it follows that . Suppose that . Applying condition (2)* with and , we obtain , in particular . Thus the only remaining option is . If or , then (2)* (with and or , respectively) again yields . Since , we are forced to conclude that . It follows that . Since , this yields . Yet now we have and (2)* says that . This also implies .
If and , then .
Hence there is with and . There is thus no loss of generality in assuming .
Now and for all of order less than . This combines with (3)* to yield:
If , then .
Since and , condition (5)* implies that and .
Suppose that . Then the current restrictions on a entail only if , in particular . But , so this is made impossible by (5)*. Accordingly, ; since is admissible whenever satisfies , we may fix notation such that . If , then every triple that satisfies is admissible and notation may again be arranged such that .
Let as before. By (3)*, we have . If satisfies , then is admissible. Thus may be assumed to be in W, whence results. By (3)*, this implies that . Since , there is with Since this means , (1)* says that and . However, and it follows that , a contradiction that ends this proof. ∎
Let P be a finite 2-group possessing a normal subgroup Q such that the following conditions are satisfied:
, where and .
, where for .
Then P is not core-2.
Let , let and let . Like in the previous proof, a triple of elements of Q will be called admissible whenever (c) continues to be true after replacing by for . Observe that (b) implies whenever and , in particular whenever . Let . Combining conditions (a) and (d), we obtain , in particular whenever . Thus and (a) implies that or . Since , we have .
Assume that P is core-2. Our first goal is showing that some admissible triple satisfies . Assuming that such a triple does not exist, we have . Upon replacing by if necessary, we may take . If , then there is with . If , then is an admissible triple endowed with the extra feature we are currently seeking. If , then is an involution in , which leaves the only undiscarded option. Yet that yields ; letting , we have with . One of the three cyclic four-subgroups of R must be normal in P, giving Q an element of breadth 1 and thus clashing with (a). Hence:
Without loss of generality, .
Observe that (1)* entails for , while and .
Since , either every element of Y is a square in Q, or is not. Assume the latter. If , then and if , then there is an involution in and condition (a) is contradicted. Accordingly, . Interchanging and and replacing by if necessary, we may assume . Since is admissible and , one of the subgroups and may be taken to be normal. Let with .
Assume that . By the previous paragraph’s results, is a square in Q or with and . Since , this implies that . Suppose that . Setting , there is an element with . Since and Q does not have elements of breadth 1, this is not possible.
We have found to imply that is a square in Q. Let and with . Then (e) says that and , which means that . Letting , we have , a contradiction. Accordingly:
Assume that . Let , and be the distinct complements of in Y. For , let denote the preimage of in V. Let . If , then because of and . Thus and , where and . On the other hand, Lemma 8 says that implies whenever . It follows that the commutator lies in the intersection of at least three complements of in Y, i.e. , contradicting (b).
We have seen that . Assume . Let and . By Lemma 8, s may be assumed to satisfy . It follows that with and . Perhaps upon replacing and by other elements of and , respectively, we may take ; bearing in mind that , we obtain that , which implies .
Let . For , we have , and , in particular . For each , is an admissible triple satisfying . We are hence free to assume for . Let . Then . Replacing V by in the preceding paragraph yields the desired contradiction.
Since , only the case is left. Then and there is . The argument is continued exactly as in the previous paragraph. ∎
3 Proof of the theorem
For the remainder of the paper, G denotes a minimal counterexample to the theorem. In particular, . In view of Lemma 3, the following definition makes sense:
Let be the set of abelian subgroups of G of maximal order subject to containing . Let . The map induces a G-isomorphism , i.e. every subgroup of is normal in . Let and let be the preimage of in G.
Note that . Also observe that , while Lemma 1 entails .
We fix an element A of and let . Let and let . The characteristic epimorphism will be denoted by a bar. Reviving earlier notation, we let .
Since every subgroup of G of order 4 intersects nontrivially, . In particular, or . We start off the proof by stating two slight, but useful, observations.
Let . Then the following hold:
Let . If , then .
Let . The assertion being certainly true if , we take . Now would imply , yet because of . Thus and Lemma 5 yields (a).
Let . Let W be a complement of in Y. Since x centralises Y, W is normalised by x. Since , conjugation by x induces an automorphism of of order 2. Since , or conjugation by x is nontrivial on . This proves (b). ∎
Suppose that . Let . Then , where , while every subgroup of D is normal in G.
Since , the subgroup has a complement C in A. Let . Select with the property that projects onto along the decomposition. Let . For , we have , and therefore possesses a complement D in A that is the direct product of cyclic factors that are normal subgroups of G. In particular, . Given in D such that for , we have . Now , while, for , we have , which shows that cannot be normal in G. Hence is. Induction on the length of d as a product of powers of the establishes the assertion. ∎
We start by establishing . By way of contradiction, assume there is such that . Lemma 1 yields the existence of satisfying whenever . Note that the definition of H implies . Applying Lemma 5 to the groups , , we obtain .
Let . By Lemma 1, , so there is with . If , then and Lemma 5, applied to , provides a contradiction. It follows that . Applying Lemma 2 yields a direct decomposition with and all subgroups of C normal in G. Lemma 1 then yields , and, according to Lemma 1, and .
If with , then Lemma 1 implies ; hence the only maximal subgroup of possibly normal in G is . Thus . Since and we may, perhaps upon replacing a by a different element of aN, assume that .
Let . Then , and, as seen above, this implies . If , then and Lemma 5 provides a contradiction when applied to the section . Hence . Thus , and the preimage of in H are one and the same – maximal – subgroup.
Let and let with . Since , there is such that . Since , we obtain , in particular . Since , it follows that is of the form for suitable integers α and β and . Thus we have . Since , there is satisfying . It follows that with . Since whenever , . Next we show . Suppose otherwise; then, since , there is with . Letting , we obtain that with .
Thus , in particular . Since , . As , with . It follows that . Pick with . If , then, since , , if , then anyway. Let and . We have just seen that , whence neither nor is contained in . It follows that none of the maximal subgroups of is normal in G.
We are done proving . Assume there is with . Then with . We apply Lemma 2 to obtain a decomposition , where and every subgroup of C is normal in G. Note that , and with . Since both and the preimage of in H are proper subgroups of H, we may additionally assume . This yields , contradicting . ∎
Let be minimal with the property that there is in which not every subgroup of is normal in .
We shall always assume A to be chosen to actually possess a cyclic subgroup of order that is not normal in H (). It may be useful to keep in mind that (see Lemma 6).
and unless .
Let . According to Lemma 3, . It follows that is an element of inverted by , which is possible only if .
Suppose that . Together with Lemma 6, the above implies that , i.e. . Thus necessitates . Suppose that . We have seen that . Lemma 6 consequently yields for . Without loss, there are positive integers m and satisfying , and . It follows that
Thus unless . Assume the latter. Let . Then . Since ,
Thus with and . We have seen this to be incompatible with . ∎
The cases and have been allocated their own subsection each. It will soon turn out to be the case that is the most difficult case by far.
3.1 The case
Assume . By Lemmas 6, 9, and 4, , , and . It follows that . Since , we have , while Lemma 1 says that whenever . Accordingly, . Let and let . Note that because of Lemma 6 and that is a normal elementary abelian 4-subgroup of G.
Assume there is . Since , either , or . In either case . Since is independent of a, it follows that and , i.e. with . Let . Since , we have and since , there is an element satisfying . Yet if , then , whence . It follows that with and . Since , this forces the conclusion , and , from which results. However, we have seen that , which, since , contradicts the fact that . Accordingly,
Assuming that , we can find . Induction yields a normal subgroup B of G with , , and . If , then , a contradiction. Hence . However, and if , then , leaving and the only possibility. Yet . Accordingly,
For a supplement Q of in H, (3.2) yields , in particular . We let and . Note that . Let and let be of order . As a cyclic subgroup of maximal order in , has a complement D in . Note that . Since , every coset with contains an element y with . Since we know that , it follows that .
Let d be an element of D of maximal order and let . Then
and there is with , in particular . Assume that . Let . Note that . Without loss is a power ; note that . It follows that . There is hence no loss in presuming . It follows that , in particular . If , then and . Accordingly, .
Finally, let . Since , there is an element satisfying . Since and , there is an element such that . Hence we may take and it follows that . Accordingly, and . Each of the elements has order at least 8, in particular . Let . We have established that P satisfies conditions (a), (b), and (c) of Lemma 11. Moreover, we know that and . Thus either or , and any element s of satisfies with . Any such s has to centralise Q, which shows that condition (d) in the premise of Lemma 11 applies as well. So Lemma 11 yields a contradiction. ∎
If , then whenever .
Next, – otherwise , i.e. and . Since the only maximal subgroup of U available for normality in G is . Note that this is possible only if .
If , then a power of satisfies . Letting , implies and , a contradiction. We summarise:
Let . Then
If , then , contradicting (1). Thus:
and holds for every .
Furthermore, (1) implies that , i.e.
Assume that . If , then (2) implies , while (3) says . Thus . Thus – otherwise – and , impossible. Accordingly, . Let , observing that . Suppose that . Then, without loss, , and , contradicting (1). If , then there is a generator of with the property that
Hence there also is a generator of satisfying
Since , it follows that is now an involution contained in , contradicting . Now is the only option left: then there is a generator of satisfying
i.e. , i.e. . Yet this implies and clashes with .
The outcome of the preceding two paragraphs is that and . If , then , while and the pair satisfies every constraint previously placed on . We summarise:
, , and u and v may be chosen to satisfy .
We shall assume from now on. By (2),
Thus, (1) yields .
Suppose that A has an element b of order 8. Then , while . Every assertion so far made about equally applies to , whence (5) delivers a contradiction. It follows that
, in particular .
Together with , (6) implies that . Let . By (1) and (5), , and (6) yields the existence of a complement of B in A, to be called D. For , ; thus D may be taken to be a direct product of subgroups with . Let be such that . Then , because . Thus for all subgroups E of D. Let . We have seen that some element w of is in . Now , i.e. . This forces the conclusion . Now and , i.e. .
, with every subgroup of D normal in G.
Assuming , let . Due to , we may replace v by vd in (1)–(4), to obtain . If , this yields , a contradiction. By (7), this means that v inverts each element of D. It follows that , and, since , we obtain . Now is impossible, as it would imply
Thus – according to (7), . Let . From , we infer that . Since both and are normal in G, this yields
Since and , we have
i.e. . Thus we obtain that , and
If , then .
Assume otherwise. Pick with and let . By Lemma 6, and by Lemma 6, so . In particular, and Lemmas 1 and 6 yield . The assumption , however, implies that . If this is true, we may assume , where and . Either , or satisfies , , and , a contradiction. If , we have ; yet and is contradicted.
Accordingly, ; by Lemma 6, this means . If , then there is such that , and forces the conclusion . Thus . Suppose there is such that . Then, as we have seen, and . It follows that is an involution with . Finally, Lemmas 1 and 6 say that whenever , implying . We summarise:
If , then . Any element v of H with has order 8 and centralises .
Let and let . For all we have , i.e. . By (1), the Hughes subgroup of is contained in , whence and every element of K is inverted by y modulo T. Suppose there is with . By (1), , whence whenever , in particular . However, , contradicting (1). Hence . Let , noting that, since , we have . If , then and . It follows that y normalises every subgroup of . Accordingly, and . Applying Lemma 3, we obtain . We summarise:
, in particular and .
From and Lemma 9, we further infer:
and either , or with . In particular, and .
Suppose there is an element . Via induction, G has a normal subgroup B of index 4 satisfying . Unless B is abelian, . Given , we have , and . It is, however, possible to choose z distinct from . Thus:
We consider the possibility that and . First suppose that , i.e. . Letting , we obtain with and . Since and “” would necessitate the existence of involutions in , and . Let . By (4), . As y inverts the elements of , . Since , . Let . We have already noted that P satisfies (a)–(c) in the premise of Lemma 11. Suppose that . It follows that and any must be of the form with . Now so that and implies . Hence P satisfies the hypothesis of Lemma 11 and is not core-2. Thus:
If and , then .
We continue to work from the assumptions and . By (5), we may write . Letting , we have and . Let . Since and , . From (1) and (2) we know that , so if , then and implies . Suppose that . Then we have . Any involution in is equal to a product with , in particular and . Letting , P therefore satisfies the hypothesis of Lemma 12 and is not core-2. Consequently:
If , then .
We stick to the hypothesis . As , it follows that whenever . By (6), , and there is with . For such a v, though, is an abelian subgroup of K containing and of greater order than A. So:
Assume that . From (2) and , , where and . By Lemma 6, implies . Note that this excludes . Indeed, if that was the case, then (4) would imply and there would be satisfying . However, this would make an involution outside .
Let and . We have seen . So set . If , then is a maximal subgroup of Y not containing and satisfying , a contradiction. This places inside . Set . Lemma 8 says that
which implies or . Thus . However, Lemma 8 also says that for at least two of the three maximal subgroups W of Y that do contain , namely , , and . This is not the case.
, , , , and .
From we deduce . Let .
If and , then either or , the latter forcing . Let with . We have seen that . Since and , it follows that and has a complement in A. Let and let b be an element of maximal order in D. Since , . Let , noting that . Each of the subgroups , and is normal in H, in particular . Let . Then . It is therefore impossible that both and are elements of . Accordingly:
Suppose that . Let b be an element of A of order 8. Letting , induction yields a normal subgroup B of G with , and . Since is abelian, . In conjunction with Lemma 1 (b), (2) implies that, for , there is with . In particular, we have for . Now implies that ; yet if , then . Hence:
Since and whenever , Lemma 3 says that:
For every , .
Suppose that and . By condition (3), holds for . Since H thus is void of elements of breadth 1, it follows that whenever . Accordingly, and whenever . It now follows from (4), from (b) in Lemma 1 and from the definition of k that and that, for , there is an element with . Let , where . Because of , ; if , then , and, since the coset is not permitted to contain involutions, . In any case there is with and . We replace v by w if necessary.
Since , it follows that , where . If , then forces . Let . By (2) and (4), we have , in particular . Let . If , then and an involution in is equal to is for some . If , then , contradicting . Accordingly, . If , then P satisfies Lemma 11 with , , , if , then P fits the criteria of Lemma 12. Thus G is not core-2. Consequently:
Suppose that . By (5), we have . By Lemma 9 and since , , and , where , , , , , and . Let ; then B is a maximal normal abelian subgroup of H, though not necessarily an element of . Lemma 9 nevertheless yields the existence of with , i.e. . However, this implies that , a contradiction. Hence
Assume that , and let and . Since , either or there is an element such that . In either case, for some . No element of A has breadth 1 in H, whence . Thus we have , a contradiction. Hence
Let a be some element of A of order 4 with . As previously noted, has a complement in A of exponent greater than 4. Let . By (6) and since , with . If , then , whence . According to Lemma 9, we may assume