On Elements of Order P in Powerful P-Groups

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Journal of Algebra 270 (2003) 1–6 www.elsevier.com/locate/jalgebra

On elements of order p in powerful p-groups

L. Héthelyi a,∗,1 and L. Lévai b,2

a Department of Algebra, Institute of Mathematics, Technical University of Budapest, M˝uegyetem rakpart 3, H-1111 Budapest, Hungary b Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences, POB 127, H-1364 Budapest, Hungary Received 21 September 2000 Communicated by George Glauberman

Abstract

We investigate the set Ω{1}(P ) of elements of order at most p in a powerful p-group P and prove p that |Ω{1}(P )|=|P : P |. As a corollary, we obtain a necessary and sufﬁcient condition for Ω1(P ) to be of exponent p. We give an example to show that for p = 2 there is a powerful 2-group such that Ω1(P ) is not of exponent 2.  2003 Elsevier Inc. All rights reserved.

Keywords: p-group; Powerful; Powers

1. Introduction In this paper we will investigate the set of elements of order at most p in a powerful p-group P . Recall that a p-group is called powerful if its maximal exponent-p (exponent- 4ifp = 2) quotient is Abelian (for a reference on powerful p-groups see [3] and [5]). To simplify notation in this paper p stands for a prime, Ω{1}(G) denotes the set of elements of order p of the ﬁnite p-group G, Gn denotes the subgroup generated by the nth powers in G and P denotes a powerful p-group. p P. Hall proved in the 30’s that in a regular p-group G the equality |Ω{1}(G)|=|G : G | holds and Ω{1}(G) forms a subgroup (see [4]). Since then several authors considered this problem for various classes of p-groups. In [1] Y. Berkovich counted the number of elements of order at most p in the Sylow p-subgroup of the symmetric group of degree pn.

* Corresponding author. E-mail addresses: [email protected] (L. Héthelyi), [email protected] (L. Lévai). 1 Research supported by grant OTKA T023437. 2 Research supported by grant OTKA F023436.

In [7] A. Mann generalized Hall’s results to semi-regular groups and some other classes of p-groups. N. Blackburn and A. Espuelas [2] have given examples showing that the number of elements of order at most p can be both smaller and bigger than the order of the maximal exponent-p quotient. Recently, A. Mann and C. Martinez have proved [8] that if in a p-group G the large majority of elements is of order at most p then the exponent of G is p. Nothing seems to be known about the set Ω{1}(P ) in a powerful p-group P . We prove the ﬁrst part of Hall’s result for powerful p-groups:

Theorem 1. Let P be a powerful p-group. For S ⊆ P let us denote by Ω{1}(S) the set of elements of S of order at most p. Then the following hold:

p (i) |Ω{1}(P )|=|P : P |; (ii) if |x|=p and N is a powerful normal subgroup of P such that x N is powerful then |Ω{1}(xN)|=|Ω{1}(N)|.

As an immediate consequence of the theorem we obtain the following.

Corollary 2. The following statements are equivalent for a powerful p-group P :

p (i) |Ω1(P )| |P : P |; p (ii) |Ω1(P )|=|P : P |; (iii) exp(Ω1(P )) = p; (iv) there exists a subgroup H of exponent p such that |H | |P : P p|.

Concerning the second part of Hall’s result we note that there exists a powerful 2-group P in which the exponent of Ω1(P ) is greater than two. Indeed, let us denote by D4 and C8 the dihedral group of order 8 and the cyclic group of order 8, respectively. It is easy to see that the central product D4 ◦ C8 of these groups is powerful but obviously exp(Ω1(D4 ◦ C8)) = 4. We do not know such examples for odd primes. We note that if Ω{1}(P ) is a subgroup for odd primes then that would give a partial answer to A. Mann’s question:

Question [6, Question 26]. Which p-groups are subgroups of powerful p-groups?

= In the remainder of this paper we set n logp(exp(P )) and π denotes the pth power map, that is, π(x) = xp for any element x of P .

2. Elements of order p and the pth power map

We start with investigating the properties of the map π. First we establish the following useful lemma. L. Héthelyi, L. Lévai / Journal of Algebra 270 (2003) 1Ð6 3

2 Lemma 3. Suppose that for some element x of P we have xp ∈ P p .ThenxPp contains an element of order at most p.

Proof. We may assume that x is not an element of P p. It is enough to prove the following: k if xp ∈ P p for some natural number k 2, then there is an element y in xPp such that k+1 k k−1 yp ∈ P p .Letxp ∈ P p . Then there exists an element u in P p such that up = xp. k+1 Consider the subgroup H of P generated by x and u.SetP = P/Pp and let bar denote homomorphic images in P .ThenH is Abelian if p = 2 and has class at most two otherwise. Thus H is regular and so H p = xp = up , in particular, |H p| p. p p Hence |H : Ω1(H)| p.However,Ω1(H) is not contained in P for |u| >p, u ∈ P , and exp(Ω1(H))= p. Thus there exists an element w of P such that w is contained in p p Ω1(H)\ P . Moreover, |H : (P ∩ H)|=p, thus H = P p ∩ H x = P p ∩ H w

k+1 follows. It is obvious that wp ∈ P p .Since|H : (P p ∩ H)|=p there is a suitable power y of w such that xPp = yPp, proving the lemma. ✷

Proposition 4. For any element x of P we have

2 (i) π(xPp) = xpP p ; −1 p p2 p (ii) π (x P ) = xΩ1(P )P .

2 Proof. To prove (i) we use induction on the exponent of P .IfP p = 1, then P is Abelian if p = 2 and its class is at most two otherwise. In any case π is an endomorphism of P with Ker π = P p. This settles (i) in this case. Let exp(P ) = pn >p2 and suppose that we have already established (i) for powerful p-groups of exponent less than pn. It is enough 2 n−1 to see that xpP p ⊆ π(xPp).SetP = P/Pp and let bar denote homomorphic images 2 in P . By the induction hypothesis we have π(xP p) = xpP p2 . Thus for any u ∈ xpP p n−1 there are elements z ∈ P p and y ∈ xPp such that yp = uz.Letw be an element of n−2 P p such that wp = z−1. The subgroup y,w is Abelian if p = 2 and has class at most two otherwise. Thus (yw)p = ypwp = (uz)z−1 = u. Hence u ∈ π(xPp) showing 2 xpP p ⊆ π(xPp). This proves (i). 2 To establish (ii) note that π : P/Pp → P p/P p is a homomorphism and its kernel 2 consists of cosets of P p containing an element x such that xp is in P p . By Lemma 3, these are exactly the cosets of P p containing an element of order at most p. This shows that the p p preimage of Ker π under the natural projection from P onto P/P is just Ω1(P )P .It −1 p p2 p also follows from the above argument that the set π (x P ) is the coset xΩ1(P )P ,as claimed. ✷

k k+1 Corollary 5. Let x ∈ P \ P p and suppose that xp ∈ P p for some positive integer k. Then the coset xPp contains an element y of order at most pk. 4 L. Héthelyi, L. Lévai / Journal of Algebra 270 (2003) 1Ð6

2 Proof. We use induction on k.Ifxp ∈ P p then xPp contains an element of order p 2 by Lemma 3. Suppose that xp ∈ P p \ P p . Then by induction there is an element u 2 in xpP p such that |u| pk−1. Since by Proposition 4(i), the pth power map π is 2 surjective from xPp to xpP p there is an element y in xPp such that yp = u. Obviously, |y|=p|u| pk . ✷

Corollary 6. The pth power map π sets up an index preserving bijection between the set p p p2 of subgroups of P containing Ω1(P )P and the set of subgroups of P containing P .

2 Proof. Let P = P/Pp .Thenπ induces an endomorphism π of P andanisomorphism p p between P/Ker π and Im π = P . By Proposition 4(ii) we have Kerπ = Ω1(P )P . p p p2 Therefore π induces an isomorphism between P/Ω1(P )P and P /P .Theresult follows. ✷

Theorem 1. Let P be a powerful p-group. For S ⊆ P let us denote by Ω{1}(S) the set of elements of S of order at most p. Then the following hold:

p (i) |Ω{1}(P )|=|P : P |; (ii) if |x| p and N is a powerful normal subgroup of P such that x N is powerful then |Ω{1}(xN)|=|Ω{1}(N)|.

p Proof. We prove (i) and (ii) together using induction on the order of P .IfΩ1(P ) P then we are done by induction. Otherwise Ω{1}(P ) can be decomposed into a disjoint union p of subsets of the form Ω{1}(xiP ) for certain elements xi ∈ P , i = 1,...,s for some s. p p The cosets xiP are precisely the cosets of P containing an element (say xi ) of order at most p. s p ⊆ p Obviously, i=1 xiP Ω1(P )P . We show that equality holds. Note that since 2 2 P/Pp is regular taking two elements a and b of order p we have (ab)p ∈ P p . It follows p p from Lemma 3 that the coset abP contains an element of order at most p. Thus Ω1(P )P is precisely the union of cosets of P p containing an element of order at most p. p Now xiP is powerful and is a proper subgroup of P . Hence by induction, part (ii) p p p holds for xi P (where xi = x and P = N). In particular, we have |Ω{1}(xiP )|= p p p p2 p |Ω{1}(P )|. Again by induction, part (i) applied to P yields |P : P |=|Ω{1}(P )|= p |Ω{1}(xiP )|. To establish part (i) for P it remains to prove that p k p p p p p2 |P : P |=p Ω{1}(P ) = Ω1(P )P : P P : P , (∗)

k p p p where p =|Ω1(P )P : P |. As we have seen above, Ω1(P )P is the union of those p cosets of P in P that contain an element of order at most p. Choose a set {y1,y2,...,yk} p p of elements of order p such that they form a basis of the vector space Ω1(P )P /P . Complete this set to minimal generating set {y1,y2,...,yk,yk+1,...,ym} of P .Since p 2 yk+1,...,ym all lie outside Ω1(P )P , these elements have order at least p .Itiswell p = = p p = = p = known that yi : i 1, 2,...,m P . Obviously, yi : i 1, 2,...,m yi : i L. Héthelyi, L. Lévai / Journal of Algebra 270 (2003) 1Ð6 5

+ + ∗ { p p p } k 1,k 2,...,m , so to prove ( ) we only need to show that the set yk+1,yk+2,...,ym p2 is linearly independent modulo P .SetQ = yk+1,yk+2,...,ym . Q is powerful since 2 Q P P p = Qp.LetP = P/Pp and let bar denote homomorphic images in P . Since P is powerful, P is regular. Note that by Lemma 3 and by the construction of Q we = p | p|=| : p| { p p p } have Ω1(Q) Q . It follows that Q Q Q , thus yk+1,yk+2,...,ym is linearly 2 2 independent modulo Qp = P p . This proves part (i) for P . Now we prove (ii) for P . By induction we may assume that N is a maximal subgroup p of P and x N = P . By the induction hypothesis, |Ω{1}(N)|=|N : N | and we have p p p seen that |Ω{1}(P )|=|P : P |. Note that since x is of order p,wehaveP = N , p p i i thus |P : P |=p|N : N |.Now|Ω{1}(x N)|=|Ω{1}((xN) )|=|Ω{1}(xN)| for any 1 i p − 1. Hence |Ω{1}(xN)|=(|Ω{1}(P )|−|Ω{1}(N)|)/(p − 1) =|Ω{1}(N)|.This ﬁnishes the proof of part (ii) and hence the theorem. ✷

The next corollary is immediate.

Corollary 2. The following statements are equivalent for a powerful p-group P :

p (i) |Ω1(P )| |P : P |; p (ii) |Ω1(P )|=|P : P |; (iii) exp(Ω1(P )) = p; (iv) there exists a subgroup H of exponent p such that |H | |P : P p|.

{ | | ∈ } Let S be a subset of P .Deﬁneω(S) to be the set logp s : s S .

2 Proposition 7. Let x be an arbitrary element of P and S a subset of P with π(S) = xpP p . Then ω(S) is an interval in the nonnegative integers. In particular, this is true for S = xPp p and S = xΩ1(P )P .

p p p p2 Proof. By Proposition 4 the sets xP and xΩ1(P )P are both mapped onto x P under π. We proceed by induction on the order of P . 2 Applying the induction hypothesis to xp, P p,andxpP p , we may assume that 2 ω(xpP p ) is an interval. Note that the orders of elements of S are the p-tuples of the order of elements of π(S) with the possible exception of 1 if the identity element of P is contained in S. Thus if ω(π(S)) ={k,k + 1,...,m} for some natural numbers k and m then ω(S) ={k + 1,k+ 2,...,m+ 1}, or if the identity element is contained in S (in this case k is zero) then ω(S) ={0, 1,...,m+ 1}. ✷

We need the following simple lemma.

Lemma 8. Let G = x,y be an arbitrary p-group such that both x and y are of order p2 and xp = yp. Suppose that G has class at most two if p is odd and G is Abelian if p = 2. Then x Ω1(G) = G = y Ω1(G). 6 L. Héthelyi, L. Lévai / Journal of Algebra 270 (2003) 1Ð6

p p Proof. Since G is regular and G = x has order p, the index of Ω1(G) is p.Theresult follows. ✷

n−1 n−2 Proposition 9. Let z ∈ P p and let w ∈ P p a non-identity element with wp = z.Ify p is a non-identity element of P with y = z then y Ω1(P ) = w Ω1(P ). In particular, if z has a pth root in Ω1(P ) then all pth roots of z are in Ω1(P ), and if we set Z = z then Ω1(P/Z) = w Ω1(P )/Z.

Proof. Clearly, we may assume that z = 1, that is, |Z|=p. Applying Lemma 8 to G = w,y , the result follows. ✷

pn−1 Proposition 10. Let Z P .Then|Ω1(P/Z)| |Ω1(P )|.IfΩ1(P ) is of exponent p then so is Ω1(P/Z).

Proof. The ﬁrst statement follows from Proposition 9 by induction on the order of Z.The second statement is now immediate from Corollary 2. ✷

Acknowledgment

The authors are grateful to the referee for the many helpful comments and suggestions.

References

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