Impartial achievement games for generating nilpotent groups

Bret J. Benesh; Dana C. Ernst; Nándor Sieben

doi:10.1515/jgth-2018-0117

Article Publicly Available

Impartial achievement games for generating nilpotent groups

Bret J. Benesh , Dana C. Ernst and Nándor Sieben

Published/Copyright: December 18, 2018

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From the journal Journal of Group Theory Volume 22 Issue 3

Abstract

We study an impartial game introduced by Anderson and Harary. The game is played by two players who alternately choose previously-unselected elements of a finite group. The first player who builds a generating set from the jointly-selected elements wins. We determine the nim-numbers of this game for finite groups of the form T×H, where T is a 2-group and H is a group of odd order. This includes all nilpotent and hence abelian groups.

1 Introduction

Anderson and Harary [2] introduced an impartial combinatorial game in which two players alternately take turns selecting previously-unselected elements of a finite group G until the group is generated by the jointly-selected elements. The first player who builds a generating set from the jointly-selected elements wins this achievement game denoted by 𝖦𝖤𝖭⁢(G). The outcome of 𝖦𝖤𝖭⁢(G) was determined for finite abelian groups in [2]. In [3], Barnes provides criteria for determining the outcome for an arbitrary finite group, and he applies his criteria to determine the outcome of some of the more familiar finite groups, including cyclic, abelian, dihedral, symmetric, and alternating groups.

A fundamental problem in game theory is to determine nim-numbers of impartial two-player games. The nim-number allows for the easy calculation of the outcome of the sum of games. A general theory of impartial games appears in [1, 13]. A framework for computing nim-numbers for 𝖦𝖤𝖭⁢(G) is developed in [9], and the authors determine the nim-numbers for 𝖦𝖤𝖭⁢(G) when G is a cyclic, abelian, or dihedral group. The nim-numbers for symmetric and alternating groups are determined in [4] while generalized dihedral groups are addressed in [6].

The task in this paper is to determine the nim-numbers of 𝖦𝖤𝖭⁢(G) for groups of the form G=T×H where T is a finite 2-group and H is a group of odd order. These groups have a Sylow 2-direct factor. Finite nilpotent groups are precisely the groups that can be written as a direct product of their Sylow subgroups, so the class of groups with a Sylow 2-direct factor contains the nilpotent groups. Note that groups with a Sylow 2-direct factor are necessarily solvable by the Feit–Thompson Theorem [10].

Anderson and Harary [2] also introduced a related avoidance game in which the player who cannot avoid building a generating set loses. As in the case of the achievement game, Barnes [3] determines the outcome for a few standard families of groups, as well as a general condition to determine the player with the winning strategy. The determination of the nim-numbers for the avoidance game for several families of groups appears in [4, 5, 9]. Similar algebraic games are studied by Brandenburg in [7].

2 Preliminaries

We now give a more precise description of the achievement game 𝖦𝖤𝖭⁢(G) played on a finite group G. We also recall some definitions and results from [9]. In this paper, the cyclic group of order n is denoted by ℤn. Other notation used throughout the paper is standard such as in [12]. The nonterminal positions of 𝖦𝖤𝖭⁢(G) are exactly the nongenerating subsets of G. A terminal position is a generating set S of G such that there is a g∈S satisfying 〈S∖{g}〉<G. The starting position is the empty set since neither player has chosen an element yet. The first player chooses x1∈G, and the designated player selects xk∈G∖{x1,…,xk-1} at the kth turn. A position Q is an option of P if Q=P∪{g} for some g∈G∖P. The set of options of P is denoted by Opt⁡(P). The player who builds a generating set from the jointly-selected elements wins the game.

It is well known that the second player has a winning strategy if and only if the nim-number of the game is 0. The only position of 𝖦𝖤𝖭⁢(G) for a trivial G is the empty set, and so the second player wins before the first player can make a move. Thus, 𝖦𝖤𝖭(G)=*0 if G is trivial. For this reason, we will assume that G is nontrivial for the remainder of this section, and we will not need to consider trivial groups until Section 4.

The set ℳ of maximal subgroups play a significant role in the game. The last two authors define in [9] the set

ℐ:={⋂𝒩:∅≠𝒩⊆ℳ}

of intersection subgroups, which is the set of all possible intersections of maximal subgroups. We also define 𝒥:=ℐ∪{G}. The smallest intersection subgroup is the Frattini subgroup Φ⁢(G) of G.

For any position P of 𝖦𝖤𝖭⁢(G) let

⌈P⌉:=⋂{I∈𝒥:P⊆I}

be the smallest element of 𝒥 containing the position P. We write ⌈P,g1,…,gn⌉ for ⌈P∪{g1,…,gn}⌉ and ⌈g1,…,gn⌉ for ⌈{g1,…,gn}⌉ if g1,…,gn∈G .

Two positions P and Q are structure equivalent if ⌈P⌉=⌈Q⌉. The structure classXI of I∈𝒥 is the equivalence class of I under this equivalence relation. Note that the definitions of ⌈P⌉ and XI differ from those given in [4, 5, 6, 9], but it is easy to see that these definitions are equivalent to the originals. We let

𝒴:={XI:I∈𝒥}.

We say XJ is an option of XI if Q∈Opt⁡(P) for some P∈XI and Q∈XJ. The set of options of XI is denoted by Opt⁡(XI).

The type of the structure class XI is the triple

type⁡(XI):=(|I|⁢ mod ⁢2,nim⁡(P),nim⁡(Q)),

where P,Q∈XI with |P| even and |Q| odd. This is well-defined by [9, Proposition 4.4]. We define the option type of XI to be the set

otype⁡(XI):={type⁡(XJ):XJ∈Opt⁡(XI)}.

We say the parity of XI is the parity of |I|.

The nim-number of the game is the nim-number of the initial position ∅, which is an even-sized subset of Φ⁢(G). Because of this, nim⁡(𝖦𝖤𝖭⁢(G)) is the second component of

type⁡(XΦ⁢(G))=(|Φ⁢(G)|⁢ mod ⁢2,nim⁡(∅),nim⁡({e})).

We use the following result of [9] as our main tool to compute nim-numbers. Note that type⁡(XG)=(|G|⁢ mod ⁢2,0,0). Recall that for a subset A⊆ℕ∪{0}, mex⁡(A) is the least nonnegative integer not in A.

Proposition 2.1.

For XI∈Y define

AI={a:(ϵ,a,b)∈otype(XI))},BI={b:(ϵ,a,b)∈otype(XI))}.

Then type⁡(XI)=(|I|⁢mod⁢ 2,a,b), where

a:=mex⁡(BI),b:=mex⁡(AI∪{a}) if |I| is even,

b:=mex⁡(AI),a:=mex⁡(BI∪{b}) if |I| is odd.

$Figure 1 Example of a calculation for type⁡(XI){\operatorname{type}(X_{I})} if Opt⁡(XI)={XJ,XK}{\operatorname{Opt}(X_{I})=\{X_{J},X_{K}\}}, where XI{X_{I}} and XJ{X_{J}} are odd and XK{X_{K}} is even. The ordered triples are the types of the structure classes.$

Figure 1

Example of a calculation for type⁡(XI) if Opt⁡(XI)={XJ,XK}, where XI and XJ are odd and XK is even. The ordered triples are the types of the structure classes.

The previous proposition implies that the type of a structure class XI is determined by the parity of XI and the types of the options of XI. Figure 1 shows an example of this calculation when XI is odd.

3 Deficiency

We will develop some general tools in this section. For a finite group G, the minimum size of a generating set is denoted by

d⁢(G):=min⁡{|S|:〈S〉=G}.

The following definition, which first appeared in [6], is closely related to d⁢(G).

Definition 3.1.

The deficiency of a subset P of a finite group G is the minimum size δG⁢(P) of a subset Q of G such that 〈P∪Q〉=G. For a structure class XI of G, we define δG⁢(XI) to be δG⁢(I).

Note that P⊆Q implies δG⁢(P)≥δG⁢(Q).

Proposition 3.2.

If S∈XI, then δG⁢(S)=δG⁢(I).

Proof.

Let n:=δG⁢(I) and m:=δG⁢(S). Since S⊆I, it follows as mentioned above that n≤m. Now let h1,…,hn∈G such that 〈I,h1,…,hn〉=G. For a maximal subgroup M, I⊆M if and only if S⊆M since S∈XI. Then since 〈I,h1,…,hn〉 is not contained in any maximal subgroup, we conclude that neither is 〈S,h1,…,hn〉. Thus, 〈S,h1,…,hn〉=G and δG⁢(S)≤δG⁢(I), and so we have δG⁢(S)=δG⁢(I). ∎

Corollary 3.3.

The deficiency of a generating set of a finite group G is 0 and δG⁢(∅)=δG⁢(Φ⁢(G))=d⁢(G).

Definition 3.4.

Let G be a finite group, ℰ be the set of even structure classes, and let 𝒪 be the set of odd structure classes in 𝒴. We define the following sets:

𝒟m:={XI∈𝒴:δG⁢(I)=m},𝒟≥m:=⋃{𝒟k:k≥m},

ℰm:=ℰ∩𝒟m,ℰ≥m:=⋃{ℰk:k≥m},

𝒪m:=𝒪∩𝒟m,𝒪≥m:=⋃{𝒪k:k≥m}.

Proposition 3.5 ([6, Proposition 3.8 and Corollary 3.9]).

Let G be a finite group and let m be a positive integer. If XI∈Dm, then XI has an option in Dm-1, and every option of XI is in Dm∪Dm-1. Moreover, if XI∈Em, then XI has an option in Em-1, and every option of XI is in Em∪Em-1.

Note that 𝒟0={XG}. Also, Proposition 3.5 implies that nim⁡(P)≠0 for all X⌈P⌉∈𝒟1. In the next lemma, we will use πi to denote the projection of a direct product to its ith factor.

Lemma 3.6.

If G and H are finite groups and S⊆G×H, then

δG×H⁢(S)≥max⁡{δG⁢(π1⁢(S)),δH⁢(π2⁢(S))}.

Proof.

Let (x1,y1),…,(xk,yk)∈G×H be such that

〈S,(x1,y1),…,(xk,yk)〉=G×H.

Then 〈π1⁢(S),x1,…,xk〉=G and 〈π2⁢(S),y1,…,yk〉=H, which yields the desired result. ∎

Lemma 3.7.

If G and H are finite groups and S⊆G, then

δG×H⁢(S×H)=δG⁢(S).

Proof.

By Lemma 3.6, we have δG×H⁢(S×H)≥δG⁢(S). Now let n:=δG⁢(S). Then there exist g1,…,gn∈G such that 〈S,g1,…,gn〉=G. Then

〈S×H,(g1,e),…,(gn,e)〉=G×H.

Thus, δG⁢(S)≥δG×H⁢(S×H). ∎

Lemma 3.8.

If G and H are finite groups, then

max⁡{d⁢(G),d⁢(H)}≤d⁢(G×H)≤d⁢(G)+d⁢(H).

Proof.

We have d⁢(G)=δG⁢({e}), so for K∈{G,H},

d⁢(G×H)=δG×H⁢({e}×{e})≥δK⁢({e})=d⁢(K)

by Lemma 3.6. Hence

max⁡{d⁢(G),d⁢(H)}≤d⁢(G×H).

Let n=d⁢(G) and m=d⁢(H), and let g1,…,gn∈G such that 〈g1,…,gn〉=G and h1,…,hm∈H such that 〈h1,…,hm〉=H. Then

〈(g1,e),…,(gn,e),(e,h1),…,(e,hm)〉=G×H,

d⁢(G×H)≤d⁢(G)+d⁢(H).∎

4 The achievement game 𝖦𝖤𝖭⁢(T×H)

We now determine the nim-number of 𝖦𝖤𝖭⁢(T×H), where T is a finite 2-group and H has odd order. We will split the analysis into different cases according to the parity of |T×H| and the value of d⁢(T×H).

If T is trivial, then T×H≅H and we can apply the following refinement of [9, Corollary 4.8].

Proposition 4.1.

If |H| is odd, then

𝖦𝖤𝖭⁢(H)={*0,if ⁢|H|=1,*2,if ⁢|H|>1⁢ and ⁢d⁢(H)∈{1,2},*1,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒.

Proof.

The case where |H|=1 was done in Section 2. We proceed by structural induction on the structure classes to show that

type⁡(XI)={(1,0,0),if ⁢XI∈𝒪0,(1,2,1),if ⁢XI∈𝒪1,(1,2,0),if ⁢XI∈𝒪2,(1,1,0),if ⁢XI∈𝒪≥3.

Every structure class in 𝒪0 is terminal, so type⁡(XI)=(1,0,0) if XI∈𝒪0. If XI∈𝒪1, we have {(1,0,0)}⊆otype⁡(XI)⊆{(1,0,0),(1,2,1)} by induction and Proposition 3.5, which implies type⁡(XI)=(1,2,1). Similarly, if XI∈𝒪2, then {(1,2,1)}⊆otype⁡(XI)⊆{(1,2,0),(1,2,1)}, and so type⁡(XI)=(1,2,0). Again, if XI∈𝒪3, then {(1,2,0)}⊆otype(XI)⊆{(1,1,0),(1,2,0))}, and hence we have type⁡(XI)=(1,1,0). Now if XI∈𝒪≥4, then otype⁡(XI)={(1,1,0)} by induction, so type⁡(XI)=(1,1,0).

Since XΦ⁢(H)∈𝒪d⁢(H) by [6, Proposition 3.7], the result follows from the fact that 𝖦𝖤𝖭⁢(H) equals the second component of type⁡(XΦ⁢(H)). ∎

If T is nontrivial, we handle four cases in increasing complexity: d⁢(T×H)=1, d⁢(T×H)≥4, d⁢(T×H)=3, and d⁢(T×H)=2.

Proposition 4.2 ([9, Corollary 6.9]).

If T is a nontrivial 2-group and H is a group of odd order such that d⁢(T×H)=1, then

𝖦𝖤𝖭⁢(T×H)={*1,if ⁢T×H≅ℤ4⁢k⁢ for some ⁢k≥1,*2,if ⁢T×H≅ℤ2,*4,if ⁢T×H≅ℤ4⁢k+2⁢ for some ⁢k≥1.

Proposition 4.3 ([6, Corollary 3.11]).

If |G| is even and d⁢(G)≥4, then

𝖦𝖤𝖭(G)=*0.

The following result will be useful in the case where d⁢(T×H)≥2.

Proposition 4.4 ([6, Proposition 3.10]).

If G is a group of even order, then

type⁡(XI)={(0,0,0),if ⁢XI∈ℰ0,(0,1,2),if ⁢XI∈ℰ1,(0,0,2),if ⁢XI∈ℰ2,(0,0,1),if ⁢XI∈ℰ≥3.

Proposition 4.5.

If T is a nontrivial 2-group and H is a group of odd order such that d⁢(T×H)=3, then GEN(T×H)=*0.

Proof.

Let g be the element the first player initially selects, so the game position is {g}∈X⌈g⌉. If X⌈g⌉∈ℰ≥2, then the second player selects the identity e and keeps the resulting game position {g,e} in X⌈g,e⌉=X⌈g⌉.

Otherwise, X⌈g⌉∈𝒪≥2, so g has odd order and can be written as g=(e,h) for some h∈H. In this case, the second player selects (t,e) for some involution t∈T. Then the resulting position {(e,h),(t,e)} is in X⌈(e,h),(t,e)⌉=X⌈(t,h)⌉∈ℰ≥2.

In both cases the position after the second move has nim-number 0 since it is in a structure class with type (0,0,2) or (0,0,1) by Proposition 4.4. Thus, the second player wins. ∎

Lastly, we consider the case where d⁢(T×H)=2. First, we handle the subcase when Φ⁢(T) is nontrivial.

Proposition 4.6.

If T is a 2-group and H is a group of odd order such that d⁢(T×H)=2 and Φ⁢(T) is nontrivial, then GEN(T×H)=*0.

Proof.

Because Φ⁢(T×H)≅Φ⁢(T)×Φ⁢(H) by [8, Theorem 2], we conclude that the order of Φ⁢(T×H) is even. Since d⁢(T×H)=2, we have XΦ⁢(T×H)∈ℰ2, so type⁡(XΦ⁢(T×H))=(0,0,2) by Proposition 4.4, and hence 𝖦𝖤𝖭(T×H)=*0. ∎

Remark 4.7.

If d⁢(T×H)=2 and Φ⁢(T) is trivial, it follows from the Burnside Basis Theorem [11, Theorem 12.2.1] that T is isomorphic to either ℤ2 or ℤ22.

Lemma 4.8.

If T a 2-group and H is a group of odd order, then

〈S,(t,h)〉=〈S,(t,e),(e,h)〉

for all subsets S of T×H, t∈T of order 2, and h∈H.

Proof.

Since (t,h)=(t,e)⁢(e,h)∈〈S,(t,h)〉, we have

〈S,(t,h)〉⊆〈S,(t,e),(e,h)〉.

Let n be the order of h. Then (t,e)=(tn,hn)=(t,h)n∈〈S,(t,h)〉 since n is odd. We also have (e,h)=(t,h)n+1∈〈S,(t,h)〉. Hence

〈S,(t,h)〉⊇〈S,(t,e),(e,h)〉.∎

Proposition 4.9.

If H is a group of odd order and d⁢(Z2×H)=2, then

𝖦𝖤𝖭(ℤ2×H)=*0.

Proof.

Since d⁢(ℤ2×H)=2, we have d⁢(H)=2. Let g:=(x,y)∈ℤ2×H be the element the first player initially selects, so the game position is {g}∈X⌈g⌉∈𝒟≥1. If X⌈g⌉∈𝒟1, then the nim-number of {g} is clearly not zero so the next player to move, which is the second player, wins.

If X⌈g⌉∈ℰ2, then the second player selects the identity element of ℤ2×H and keeps the resulting game position {g,e} in X⌈g,e⌉=X⌈g⌉. By Proposition 4.4, type⁡(X⌈g⌉)=(0,0,2). So the second player wins since the nim-number of {g,e} is 0.

It remains to consider the case when X⌈g⌉∈𝒪2, and hence g=(0,y). In this case, the second player picks (1,e)∈ℤ2×H. We show that the resulting game position P:={(0,y),(1,e)} is in X⌈P⌉∈ℰ2. This will prove that the second player wins since again P=*0 by Proposition 4.4.

For a contradiction, assume that X⌈P⌉∈ℰ1, so 〈(0,y),(1,e),(u,v)〉=ℤ2×H for some (u,v)∈ℤ2×H. If u=0, then by Lemma 4.8,

ℤ2×H=〈(0,y),(1,e),(0,v)〉=〈(0,y),(1,v)〉.

If u=1, then we claim that

ℤ2×H=〈(0,y),(1,e),(1,v)〉=〈(0,y),(1,v)〉.

Clearly, 〈(0,y),(1,v)〉⊆〈(0,y),(1,e),(1,v)〉, and (1,e)∈〈(0,y),(1,v)〉 by Lemma 4.8, so 〈(0,y),(1,e),(1,v)〉⊆〈(0,y),(1,v)〉. Thus, the claim holds. In either case, there is an h∈ℤ2×H such that 〈g,h〉=ℤ2×H. This implies that X⌈g⌉∈𝒪1, which contradicts the assumption that X⌈g⌉∈𝒪2. Thus, we must have X⌈P⌉∈ℰ2. ∎

Proposition 4.10.

If H is a group of odd order such that d⁢(H)≤1, then

𝖦𝖤𝖭(ℤ22×H)=*1.

Proof.

Since d⁢(H)≤1, it follows that ℤ22×H is abelian and we conclude that 𝖦𝖤𝖭(ℤ22×H)=*1 by [9, Corollary 8.16]. ∎

Proposition 4.11.

If H is a group of odd order such that d⁢(H)=2, then

𝖦𝖤𝖭(ℤ22×H)=*1.

Proof.

Let G=ℤ22×H. We have d⁢(G)=d⁢(H)=2 since ℤ22 and H have coprime orders. Hence 𝒟≥3=∅. Let

𝒪2a:={XI∈𝒪2:Opt⁡(XI)∩ℰ2=∅},𝒪2b:=𝒪2∖𝒪2a.

We will show that 𝒪1=∅, and that ℰm for m∈{0,1,2}, 𝒪2a, and 𝒪2b are nonempty. Then we will use structural induction on the structure classes to show that

(4.1)type⁡(XI)={(0,0,0),if ⁢XI∈ℰ0,(0,1,2),if ⁢XI∈ℰ1,(0,0,2),if ⁢XI∈ℰ2,(1,1,0),if ⁢XI∈𝒪2a,(1,1,2),if ⁢XI∈𝒪2b,

as shown in Figure 2.

$Figure 2 Structure classes for 𝖦𝖤𝖭(ℤ22×H)=*1{\text{\sf GEN}(\mathbb{Z}_{2}^{2}\times H)=*1} with d⁢(H)=2{d(H)=2}.$

Figure 2

Structure classes for 𝖦𝖤𝖭(ℤ22×H)=*1 with d⁢(H)=2.

First, we show that 𝒪1 is empty. Assume L is an intersection subgroup of odd order. Then L={e}×K for some subgroup K of H. Since δℤ22⁢({e})=2, we see that δG⁢(L)≥2 by Lemma 3.6. Hence XL∉𝒪1, and we conclude that 𝒪1=∅.

Now, we show that ℰm is nonempty for m∈{0,1,2}. Let t be a nontrivial element of ℤ22 and consider K:=⌈(t,e)⌉, which has even order. Since (t,e) is contained in the maximal subgroups 〈t〉×H and ℤ22×M for every maximal subgroup M of H, it follows that K is a subgroup of 〈t〉×Φ⁢(H). Then

2=d⁢(G)≥δG⁢(K)≥δG⁢(〈t〉×Φ⁢(H))≥δH⁢(Φ⁢(H))=d⁢(H)=2,

by Lemma 3.6 and [6, Corollary 3.3]. Thus, XK∈ℰ2. Since ℰ2 is nonempty, we can conclude that ℰ1 and ℰ0 are nonempty by repeated use of Proposition 3.5. By Proposition 4.4, the types of structure classes in ℰm for m∈{0,1,2} are as described in equation (4.1).

We now show that 𝒪2a≠∅. If u∈ℤ22 is nontrivial with t≠u, then 〈t〉×H and 〈u〉×H are both maximal subgroups of G whose intersection is {e}×H. Hence {e}×H is an intersection subgroup of G with odd order. Any intersection subgroup I properly containing {e}×H must be isomorphic to ℤ2×H, so XI∈ℰ1 by Lemma 3.7. Thus X{e}×H∈𝒪2a since 𝒪1=∅.

Next, we show that 𝒪2b≠∅. By [8, Theorem 2],

Φ⁢(G)=Φ⁢(ℤ22)×Φ⁢(H)={e}×Φ⁢(H),

so Φ⁢(G) has odd order. Hence, XΦ⁢(G)∈𝒪d⁢(G)=𝒪2 by Corollary 3.3. Then

2≥δG⁢(Φ⁢(G)∪{t})≥δG⁢(ℤ22×Φ⁢(H))=δH⁢(Φ⁢(H))=d⁢(H)=2

by Lemma 3.7 and Corollary 3.3. So X⌈Φ⁢(G),t⌉∈ℰ2 by Proposition 3.2. Thus, X⌈Φ⁢(G),t⌉∈ℰ2 is an option of XΦ⁢(G), so XΦ⁢(G)∈𝒪2b.

It remains to show that

type⁡(XI)={(1,1,0),if XI∈𝒪2a,(1,1,2),if XI∈𝒪2b.

If XI∈𝒪2, then XI must have an option in ℰ1 by Proposition 3.5 since 𝒪1=∅, and so (0,1,2)∈otype⁡(XI).

Let XI∈𝒪2a. We first show that XI has no option in 𝒪2b. Suppose toward a contradiction that XJ∈𝒪2b is an option of XI, and let XJ have an option XK∈ℰ2. Let v∈K such that v has order 2. Then ⌈I,v⌉≤⌈J,v⌉≤K, so XI has an option X⌈I,v⌉∈ℰ2, which contradicts the definition of 𝒪2a. Thus, otype⁡(XI) is either {(0,1,2)} or {(0,1,2),(1,1,0)} by induction, so type⁡(XI)=(1,1,0).

Finally, let XI∈𝒪2b. Then XI has an option in ℰ2 by the definition of 𝒪2b. We will show that XI also has an option in 𝒪2a. Let h1,h2∈H such that H=〈h1,h2〉, and let J:=⌈I,(e,h1)⌉. We will show that XJ∈𝒪2a by showing that XJ∈𝒪2 and X⌈I,(e,h1),(s,x)⌉∉ℰ2 for all (s,x)∈G. Since I∪{(e,h1)}⊆{e}×H and {e}×H is an intersection subgroup of odd order, we must have J≤{e}×H. Hence XJ∈𝒪, which implies that XJ∈𝒪2 since 𝒪1=∅.

Now let (s,x)∈G. We will prove that X⌈J,(s,x)⌉∉ℰ2. If (s,x) has odd order, then s=e, so 〈I,(e,h1),(s,x)〉≤{e}×H, and thus X⌈J,(s,x)⌉∈𝒪2≠ℰ2. Thus, we may assume that s is nontrivial, and we let w∈ℤ22 be such that 〈s,w〉=ℤ22. Then

〈I,(e,h1),(s,x),(w,h2)〉=〈I,(e,h1),(e,h2),(e,x),(s,e),(w,e)〉=G

by two applications of Lemma 4.8, which implies X⌈I,(e,h1),(s,x)⌉∈ℰ1≠ℰ2. Hence XJ∈𝒪2a. Thus,

{(0,1,2),(0,0,2),(1,1,0)}⊆otype⁡(XI)

⊆{(0,1,2),(0,0,2),(1,1,0),(1,1,2)},

and so type⁡(XI)=(1,1,2). ∎

The results in this section lead to our main theorem.

Theorem 4.12.

If G=T×H, where T is a 2-group and H is a group of odd order, then

𝖦𝖤𝖭⁢(G)={*1,if ⁢|G|⁢ is odd and ⁢d⁢(G)≥3,*1,if ⁢G≅ℤ4⁢k⁢ for some ⁢k,*1,if ⁢G≅ℤ22×H⁢ with ⁢d⁢(H)≤2,*2,if ⁢G≅ℤ2,*2,if ⁢|G|⁢ is odd and ⁢d⁢(G)∈{1,2},*4,if ⁢G≅ℤ4⁢k+2⁢ for some ⁢k≥1,*0,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒.

Proof.

Each case of the statement follows from an earlier result we proved. The following outline shows the case analysis:

|G| is odd (Proposition 4.1).
|G| is even,
1. d⁢(G)=1 (Proposition 4.2),
2. d⁢(G)≥4 (Proposition 4.3),
3. d⁢(G)=3 (Proposition 4.5),
4. d⁢(G)=2,
  1. Φ⁢(T) is nontrivial (Proposition 4.6),
  2. Φ⁢(T) is trivial,
    T≅ℤ2 (Proposition 4.9),
    T≅ℤ22,
    d⁢(H)≤1 (Proposition 4.10),
    d⁢(H)=2 (Proposition 4.11).

The two cases for when Φ⁢(T) is trivial are justified by Remark 4.7. ∎

Recall that every nilpotent group, and hence every abelian group, can be written in the form T×H, where T is a finite 2-group T and H is a group of odd order. As a consequence, Theorem 4.12 provides a complete classification of the possible nim-values for achievement games played on nilpotent groups. Moreover, Theorem 4.12 is a generalization of [9, Corollary 8.16], which handles abelian groups only. Note that even in the case when H is not nilpotent, H must be solvable by the Feit–Thompson Theorem [10].

Example 4.13.

The smallest non-nilpotent group that has a Sylow 2-direct factor is isomorphic to ℤ2×(ℤ7⋊ℤ3), which has order 42.

Example 4.14.

The smallest group that does not have a Sylow 2-direct factor is S3. That is, S3 is the smallest group not covered by Theorem 4.12. However, the possible nim-values for achievement and avoidance games played on symmetric groups were completely classified in [4]. The dihedral groups Dn for n≥3 are not covered by Theorem 4.12 either, but these groups were analyzed in [9].

5 Further questions

We mention a few open problems.

What are the nim-numbers of non-nilpotent solvable groups of even order that do not have a Sylow 2-direct factor?
The smallest group G for which nim⁡(𝖦𝖤𝖭⁢(G)) has not been determined by results in [4, 5, 6, 9] or Theorem 4.12 is the dicyclic group ℤ3⋊ℤ4. All dicyclic groups have Frattini subgroups of even order. Hence these groups have nim-number 0 as a consequence of Proposition 4.4. The smallest group not covered in the current literature is ℤ3×S3. What are the nim-numbers for groups of the form ℤm×Sn for m≥2 and n≥3?
The nim-numbers of some families of nonsolvable groups were determined in [4]. Can we determine the nim-numbers for all nonsolvable groups?

Communicated by Nigel Boston

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Received: 2018-05-16

Revised: 2018-09-11

Published Online: 2018-12-18

Published in Print: 2019-05-01

Articles in the same Issue

https://doi.org/10.1515/jgth-2018-0117