ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 627-636 https://doi.org/10.26493/1855-3974.1921.d6f (Also available at http://amc-journal.eu) The symmetric genus spectrum of abelian groups Received 25 January 2019, accepted 6 November 2019, published online 13 December 2019 Let S denote the set of positive integers that appear as the symmetric genus of a finite abelian group and let S0 denote the set of positive integers that appear as the strong symmetric genus of a finite abelian group. The main theorem of this paper is that S = S0. As a result, we obtain a set of necessary and sufficient conditions for an integer g to belong to S. This also shows that S has an asymptotic density and that it is approximately 0.3284. Keywords: Symmetric genus, strong symmetric genus, Riemann surface, abelian groups, genus spectrum, density. Math. Subj. Class.: 57M60, 30F35, 20F38 1 Introduction Let G be a finite group. Among the various genus parameters associated with G, one of the most important is the symmetric genus 1. Our approach utilizes the strong constraints on the Sylow 2-subgroup of a group (not necessarily abelian) acting on a surface of even genus or a surface of genus congruent to 3 (mod 4); here see [7, Theorem 8] and [8, Theorem 5]. In the exceptional case (in which the Sylow 2-subgroup has a special form), we show that there exists an abelian group Ai such that a(A) = a0(Ax). To establish the reverse containment S0 c S, we utilize the characterization of the integers in the spectrum S0 in [3, Theorem 1]. For each integer g satisfying one of the five conditions in that result, we exhibit an abelian group G such that g = a(G). 2 Background results Let A be a non-trivial finite abelian group of rank r. Then A has the standard canonical representation A — Zmi x Zm2 x • • • x Zmr, (2.1) with invariants m1; m2,..., mr subject to m1 > 1 and m; | mi+1 for 1 < i < r. The abelian group A also has another canonical form that is useful in calculating genus parameters of abelian groups. Define the alternate canonical form for A as the direct product of three subgroups T, B and D of A. First, the group D is the subgroup of A C. L. May and J. Zimmerman: The symmetric genus spectrum ofabelian groups 629 generated by the factors Zms, where ms is divisible by 4. Then write A = D x D. Now let T be the Sylow 2-subgroup of D, which is elementary abelian, and B be its direct summand of odd order. Therefore, A = T x B x D. Define t = rank(T), b = rank(B), d = rank(D). It follows that r = rank(A) = d+max(b, t). We point out that this notation differs from that used in [5]. The groups of symmetric genus zero are the classical, well-known groups that act on the Riemann sphere (possibly reversing orientation) [2, §6.3.2]. The abelian group A has symmetric genus zero if and only if A is Zn, Z2 x Z2n, or (Z2)3; see [2, §6.3.2]. The groups of symmetric genus one have also been classified, at least in a sense. These groups act on the torus and fall into 17 classes, corresponding to quotients of the 17 Euclidean space groups [2, §6.3.3]. Each class is characterized by a presentation, typically a partial one. The abelian group A has symmetric genus one if and only if A is Zm x Zmn with m > 3, Z2 x Z2 x Z2n with n > 2, or (Z2)4; see [2, §6.3.3]. Let A be a finite abelian group. The strong symmetric genus of A has been completely determined by Maclachlan [4, Theorem 4], and if A has odd order, then a(A) = a0 (A). The focus of [5] was the determination of the symmetric genus of an abelian group of even order. The approach was to show that, among the various genus actions of A, there is one induced by an NEC group with a signature of one of three types. We established the following result [5, Theorem 3.10]. Theorem A. Let A be an abelian group of even order. Among the NEC groups with minimal non-euclidean area that act on A, there is a group r whose signature has one of the following forms: (I) (g, +, [Ai,...,An], {}); (II) (0, +, [Ai,...,As], {()k }) for some k > 1; (III) (0, +, [], {( )u, (2v)}) for some v > 2. Furthermore, in cases (I) and (II), Aj divides Ai+1 for 1 < i < r — 1. In (III) the notation (2v) means, as usual, a period cycle with v link periods equal to 2. We denote by t(A) (here and in [5]) the minimum genus of any action of A induced by an NEC group of Type II. The size of the largest elementary abelian 2-group factor of A determines whether a (A) is given by an action induced by a group of Type I, II or III. The main result of [5] is the following [5, Theorem 5.7]. Theorem B. Let A be an abelian group of even order with canonical form A = (Z2)a x Zmi x Zm2 x • • • x Zmq , where m1 > 2. If the symmetric genus a(A) > 2, then (i) a(A) = 1 + |A| • (a + 3q — 4)/8, if a > q + 2; (ii) a(A) = t(A), if 1 < a < q + 1; (iii) a(A) = min{a0(A), t (A)}, if a = 0. Thus Theorem B gives the symmetric genus of an abelian group A in terms of the invariants of A and the numbers a0 (A) and t(A). The main result in [3] is the characterization [3, Theorem 1] of the integers in the spectrum S0, and this will be important here. 630 Ars Math. Contemp. 17 (2019) 493-514 Theorem C. Let g > 2. Then g € S0 if and only if g satisfies one of the following conditions: (i) g = 1 (mod 4) or g = 55 (mod 81); (ii) g — 1 is divisible by p4 for some odd prime p; (iii) g — 1 is divisible by a2 for some odd integer a with (a — 1) | g; (iv) g — 1 is divisible by b2a2(a — 1) for some odd integers a,b> 1,with a = 3 (mod 4). 3 So CS To establish the containment S0 C S, we use the characterization of the integers in the spectrum S0 in Theorem C. For each integer g satisfying one of the five conditions in that result, we exhibit an abelian group G such that g = a(G). This is quite easy, as we shall see. In this section, we will assume that A is always written in alternate canonical form, A = T x B x D. We begin by noting some consequences of Theorem B. Directly from part (i) we have the following; this formula was also pointed out in [7, p. 4094]. Proposition 3.1. a(Zf x Z2m) = 1 + 4m for any integer m > 2. Since a((Z2)4) = 1 and a(Z)5) = 5 [7] (the general genus formula is a((Z2)n) = 1 + 2n-3(n — 4) [5, Corollary 5.4]), it follows that the spectrum S contains the entire congruence class g = 1 (mod 4). These odd integers are also in S0 [3, p. 342]. A special case of Theorem B [5, p. 423] will be important here. Theorem 3.2. Let the abelian group A have alternate canonical form A = T x B x D. If T is trivial, then a(A) = a0(A). Let A be a finite abelian group. Then a(A) = a0(A) in another important case. Lemma 3.3. Let A be an abelian group of rank three or more. If the Sylow 2-subgroup S2 of A is cyclic, then a(A) = a0(A). Proof. By Theorem 3.2, we may assume that T is non-trivial. If S2 is cyclic, then we must have S2 = Z2 and D = 1. Now write A = Z2 x B = Z2 x x Zp2 x • • • x Zpb for b > 3, where each ^ is odd. In this case, t = 1, d = 0 and b > 3. Since t < d + 1, [5, p. 416] gives . =1+2|a| (-1+1 (i - s)) (see also (4.1) in Section 4). The group A has canonical form A = Zp1 x Zp2 x • • • x Z2pb, and is the image of a Fuchsian group r with signature (0, +, ..., 2ftb], { }). When we calculate the genus arising from the action of this Fuchsian group on A, we get that it is equal to t(A). Now applying Maclachlan's formula shows that ct°(A) < t(A) and hence 5 and set G = Zp x Zp x Zp x Zmp. Then g = 1, with a = 3 (mod 4). Then g = a°(G), where G is a group of the form Za x Zab x Zabn [3, p. 343], a group with a cyclic Sylow 2-subgroup, so that 1 (the integer k is the number of empty period cycles). Among the signatures of the NEC groups that induce the Type II genus t(A), Lemmas 4.2 and 4.3 of [5] identify one value of k that can be used to calculate t(A). These two lemmas are correct. Unfortunately, there is a mistake in [5, Formula (4.5)], which is used in the final determination of t(A). We correct that here. Let A = T xB xD be the alternate canonical form for A. Remember that t = rank(T), b = rank(B), d = rank(D) and so r = rank(A) = d + max(b, t). The odd order group B is generated by elements with orders ft^ ..., ftb, where p; divides Pj for i < j. In the case in which t < d + 1, the formula for t(A) [5, p. 416] is correct. Note that k = t gives minimal area by [5, Lemma 4.2]. In the new notation, this formula is T(A) = 1 + 5|A|(k - 2 +1 0 - s) + (' - j;)). <4-1> where the group D is generated by elements with orders J1,..., Sd satisfying j divides Jj for i < j . Next, we consider the case t > d + 1. Theorem 4.1. Let A be an abelian group in alternate canonical form. Suppose that t > d +1 and k is given by [5, Lemma 4.3]. There are two cases and in each case, define v = b + d - k +1. (i) Suppose that t + d is odd. Then k = (t + d + 1)/2; (a) If b < (k - 1) - d, then t(A) = 1 + 2 |A|(k - 2); 632 Ars Math. Contemp. 17 (2019) 493-514 (b) If b > (k - 1) - d, then v > 1 and - (k - 1) - d, then v > 1 and t (A) = 1 + 2'A|(k - 2+g(1 - S) +(1 - i)). Proof. Among the NEC groups that induce the Type II genus t(A), let r be one with signature (0; +; [Ai,...; As]; ( )k) in which the number k of empty period cycles is given by k = [(t + d + 1)/2] [5, Lemma 4.3, p. 414]. It follows that k > d + 1. The group r has generators xi,..., xs, ei,..., ek, and involutions ci,..., ck. The defining relations for r consist of xi • • • xsei • • • ek = 1, conditions on the order of the elements xj and certain elements commuting. Clearly, one generator is redundant, and, since ^(r) is minimal, we may assume that ek is that generator. Now let ai,..., ar be a generating set for A in canonical form so that the orders of these elements satisfy the standard divisibility condition. With ^(r) minimal, the elements ei,..., ek-i are mapped onto the subgroup generated by the k -1 elements ar_k+2,..., ar of highest order. In particular, since k -1 > d, the subgroup D of A is contained in the image of (ei,..., ek-i). If t + d is odd, then 2k - 1 = t + d. Since t + d is the rank of the Sylow 2-subgroup S2 of A, we have that S2 is contained in the image of (ci,..., ck, ei,..., ek-i). It follows that (T, D) is contained in the image of (ci,..., ck, ei,..., ek-i). If t + d is even, then 2k = t + d. In this case, there is an additional generator x^ so that (T, D) is contained in the image of (x^, ci,..., ck, ei,..., ek-i). If b < (k - 1) - d, then the images of the generators e» which are not mapped into D generate all of B. Therefore, if t + d is odd, then A is the image of the NEC group with signature (0; +; [ ]; {( )k}) and t(A) = 1 + 2|A|(k - 2). If t + d is even, then we need a generator xi in order to map onto A, and with ^(r) minimal, |xi| = 2. Therefore, if t + d is even, then A is the image of the NEC group with signature (0; +; [2]; {( )k}) and t(A) = 1 + i|A|(k - 2). The last case is when b > (k - 1) - d. Let E be the subgroup of A generated by the images of ei,..., ek-i. Then the subgroup B can be decomposed as B = Bi x B2, where B2 = B n E. Let v = b + d - k + 1 so that v is the rank of Bi and v > 1. We need generators xi,..., xv to map onto Bi. If t + d is odd, then A is the image of the NEC group with signature (0; +; [Si,..., ]; {()k}) and t(a)=1+2|a| (k - 2+g(1 - Si ^. Suppose that t + d is even. As in the previous case, we need generators xi,..., xv to map onto Bi. However, the generator xv must map onto an element of order for the NEC group to map onto A. This is because there is the extra involution not in the image C. L. May and J. Zimmerman: The symmetric genus spectrum ofabelian groups 633 of (ci,..., ck). If t + d is even, then A is the image of the NEC group with signature (0;+;k8i,...,&-i, ]; {()k }) and T(A) = 1 + 5|A|(k - 2 + £ (' - s) + (' - n 5 General results Again in this section, we assume that A is written in alternate canonical form, A = T x B x D. Let A be a finite abelian group so that the integer g = a(A) is in the spectrum S. We want to show that g is in S0 as well. This is clearly the case if A has rank one or two or A has a trivial factor T in its alternate canonical form. Thus we may assume that A has rank at least 3 and T is not trivial. In particular, A has even order. Our approach utilizes the Sylow 2-subgroup S2 of A. Lemma 5.1. Let A be an abelian group of rank three or more with a(A) > 2. If the Sylow 2-subgroup of A is isomorphic to (Z2)3, then t (A) = 1 (mod 4). Proof. Since S2 = (Z2)3, we have T = S2, D = 1 and A = T x B. Now t = 3, d = 0, and k = 2. Since b = 1 would imply a (A) = 1, b > 2. We have t > d +1 with t + d odd. The Type II genus is t(A) = 1 + |A| • M/2, where M = (k - 2 + EV=1(1 - )) by Theorem 4.1(i)(b). Since M • |B| is an integer and |A| = 8|B|, we clearly have t(A) = 1 (mod 4). □ Lemma 5.2. Let A be an abelian group of rank three or more with a(A) > 2. If the Sylow 2-subgroup of A is isomorphic to Z2 x Z2 x Z2e for some £ > 2, then t (A) = 1 (mod 4). Proof. In this case we have T = Z2 x Z2 and D is cyclic with order divisible by 2£. Therefore, t = 2 and d =1. This implies that k = t = 2 by [5, Lemma 4.2]. Since b =1 would imply a(A) = 1, b > 2. We have t = d +1 and by (4.1), the Type II genus t(A) = 1 + |A| • M/2 where M = (k - 2 + EbU(1 - ^)). Since M • |B| is an integer and |A| =4 • |D| • |B| with |D| divisible by 2£, again t(A) = 1 (mod 4). □ Lemma 5.3. Let A be an abelian group. If the Sylow 2-subgroup of A is isomorphic to (Z2)4, then t(A) = 1 (mod 4). Proof. Since S2 = (Z2)4, we have T = S2, D = 1 and A = T x B. Now t = 4, d = 0, k = 2 and b > 1. We have t > d +1 with t + d even. We apply Theorem 4.1(ii) and get that k = 2. If b =1, then M = 1/2 by Theorem 4.1(ii)(a). If b > 2, then by Theorem 4.1(ii)(b), M • 2|B| is an integer. In either case, 2M • |B| is an integer. Since |A| = 16|B|, we again have t (A) = 1 (mod 4). □ Theorem 5.4. Let A be an abelian group of rank three or more with a(A) > 2. Then either a(A) = 1 (mod 4) or a(A) = a0 (A) (or both), unless the Sylow 2-subgroup of A has rank 2 and is isomorphic to Z2 x Z2c ,for some £ > 1. Proof. First, if A has odd order, then a (A) = a0 (A). Assume, then, that A has even order so that a (A) is given by Theorem B. Let A have canonical form A = (Z2)0 x Zmi x Zm2 x • • • x Zmq , 634 Ars Math. Contemp. 17 (2019) 493-514 as in Theorem B. It is easy to see in case (i), we always have that a(A) = 1 (mod 4). Suppose a < q + 1. By Theorem B, a(A) is either equal to a0 (A) or t(A). Let A act on a surface X of genus g = a(A) > 2, and write |A| = 2n • m, where m is odd. Assume first that g is even. Then by [7, Theorem 9], A is not a 2-group so that m = 1. We consider the possibilities for the Sylow 2-subgroup S2 of A. If S2 is cyclic, then by a (A) = a0 (A) by Lemma 3.3. Assume then that S2 is not cyclic. If A contains an element of order 2n-1 with |S21 = 2n, then S2 is isomorphic to Z2 x Z2n-i, the exceptional case. Assume then that A has no elements of order 2n-1, and apply [7, Theorem 8]. Since A and S2 are abelian, the only possibility is that S2 is isomorphic to (Z2)3. But in this case t(A) = 1 (mod 4) by Lemma 5.1. Therefore, by Theorem B, if g is even, then either a(A) = 1 (mod 4) or a(A) = a0(A), unless S2 = Z2 x Z2c, for some i > 1. Now suppose g = 3 (mod 4), and use [8, Theorem 5]. The Sylow 2-subgroup S2 also acts of the surface X of genus g > 2. By [8, Theorem 5], S2 contains an element of order 2n-3 or larger. Further, if Exp(S2) = 2n-3, then S2 contains a dihedral subgroup of index 4. We consider the possibilities for Exp(S2). If S2 is cyclic, then by a(A) = a0(A) by Lemma 3.3. If S2 is not cyclic and contains an element of order 2n-1, then S2 is isomorphic to Z2 x Z2n-i, the exceptional case. Suppose Exp(S2) = 2n-2. Then S2 is isomorphic to either Z2 x Z2 x Z2n-2 or x Z2n-2. If S2 = Z2 x Z2 x Z2n-2, then by Lemma 5.2, t(A) = 1 (mod 4). If on the other hand, S2 = Z4 x ^2^-2, then by Theorem 3.2, a(A) = a0(A). Suppose Exp(S2) = 2n-3 and S2 has a dihedral subgroup of index 4. Since S2 is abelian, this forces n = 4 and S2 = (Z2)4. In this case, t(A) = 1 (mod 4) by Lemma 5.3. Therefore, by Theorem B, if g = 3 (mod 4), then either a(A) = t(A) = 1 (mod 4) or a(A) = a0(A), unless S2 = Z2 x Z2t, for some i > 1. □ A consequence of the proof is perhaps worth noting, in connection with the well-known conjecture that "almost all" groups are 2-groups. Theorem 5.5. If A be an abelian 2-group of positive symmetric genus, then a(A) = 1 (mod 4). Proof. Assume A is an abelian 2-group with a(A) > 2. Then a(A) is not even by [7, Theorem 9]. The proof of Theorem 5.4 shows that a (A) cannot be congruent to 3 (mod 4) either, since A = S2 and A is not an abelian group of genus zero or one. □ Next we handle the exceptional case in Theorem 5.4. Theorem 5.6. Let A be an abelian group of rank three or more with a(A) > 2. If the Sylow 2-subgroup of A is isomorphic to Z2 x Z2c, for some i > 1, then there exists an abelian group A1 such that a(A) = a0(A1). Proof. Let A have alternate canonical form A = T x B x D. We consider two cases, i =1 and i > 2. C. L. May and J. Zimmerman: The symmetric genus spectrum ofabelian groups 635 First assume that S2 = (Z2)2. Now T = S2, D = 1 and A = T x B, with t = 2, d = 0, and k =1. Write A = Z2 x Z2 x B = Z2 x Z2 x Z^ x Z^2 x • • • x Z^h for b > 3, where each i is odd. By Theorem 4.1(ii)(b), the Type II genus is T (A) = 1 + 2lAl(-1 + g (1 - i ) + (1 - 2P;)). By Theorem B, a(A) = min{a0(A),r(A)}. If a(A) = ct°(A), then set Ai = A and we are done. So we assume that t(A) < 1 to obtain 1 as in the calculation of a°(A). By assumption the Fuchsian groups rp for p > 1 give genus larger than t(A). The Fuchsian group r° gives the same genus as t(A). Therefore, t(A) = ct°(A1) and so 2. Now T = Z2 and D is isomorphic to Zm2£, where m is odd. We have alternate canonical form A = Z2 x B x Zm2c = Z2 x Z^1 x Z^2 x • • • x Z^t-1 x Zm2£, where each i is odd, i divides i for i < j and m is divisible by i6_1. It follows that t = 1, b > 3, d =1, and k =1 by [5, Lemma 4.2]. By (4.1) the Type II genus is '