
SUM AND DIFFERENCE SETS IN GENERALIZED DIHEDRAL GROUPS 5
If nis arbitrarily large and jis a constant compared to n, we can make a stronger statement:
we can replace cjin the above theorem with a constant arbitrarily close to q2/7≈0.5345.
Specifically, we have the following.
Theorem 3.1. For fixed jand > 0, there exists nj, with the following property. Let Gbe an
abelian group of size n≥nj, with at most jelements of order 2or 1. Then with Dand SD,m
defined as in Theorem 1.8, we have that if 6≤m≤q2/7−√n, then there are more MSTD
sets than MDTS sets in SD,m.
We can give a stronger, more general statement on the proportion of MSTD sets in SD,m if m
is large and also bounded above by a (smaller) constant times √n.
Theorem 3.2. Let D, j, and SD,m be defined as in Theorem 1.8. For any > 0, there exist m
and c,j such that if m≤m≤c,j√n, the proportion of MSTD sets in SD,m is at least 1−.
Here mand c,j are independent of n, but similarly to before, this theorem is only useful when
n≥(m/c,j)2.
Remark 2.In practice, Theorems 1.8 and 3.2 are most useful if jis essentially a constant compared
to n. This is indeed the case for the original dihedral group D2n=Z2n Zn, where j= 1 when
nis odd and j= 2 when nis even, yielding cj≥0.12. The family of original dihedral groups is
also a good example of how to use Theorem 3.1: we can apply that theorem with j= 2 and
arbitrarily small to get that for large enough dihedral groups, we can get the coefficient of the √n
in the theorem to be very close to q2/7, which is a significant improvement over 0.12.
However, these theorems are not useful when, for example, G=Z2×. . . ×Z2×Z3. Here G
does have an element of order at least 3, so Conjecture 1.6 applies, but we have j=n/3, which
is too large for Theorems 1.8 and 3.2 to apply to any values of m. In fact, if j≥n/100, then
n≤(6/cj)2, and Theorem 1.8 does not apply to any values of m.
We first prove Theorem 1.8 here. Then, in Subection 3.1 we demonstrate Theorems 3.1 and 3.2.
Recall the definitions of RDand FDfrom Section 2.2. Note that any element in FDhas order
2 in D. Furthermore, any element in RDhas order at most 2 in Dif and only if it has order at
most 2 in G.
We begin with a set Awith size mand count the number of elements in A+Aand A−A. In
this count, we will make a naive assumption: there are no overlaps between sums and differences
that we do not expect to overlap. Decompose Ainto the union of the set of rotations R=A∩RD
of and the set of flips F=A∩FD, and define k=|F|. We have:
A+A= (R+F)∪(F+R)∪(R+R)∪(F+F);
A−A= (R−F)∪(F−R)∪(R−R)∪(F−F) (5)
Consider first the flips in A+Aand A−A. In A+A, these are in R+Fand F+R. Note
that we do not expect a lot of overlap in general; for a rotation (0, g1)∈Rand a flip (1, g2)∈F,
(0, g1)·(1, g2) = (1, g1g2) does not equal (1, g2)·(0, g1) = (1, g2g−1
1) unless g1has order 1 or 2
in G. On the other hand, for the flips in A−A, we have R−F=F−R. This is because
(0, g1)·(1, g2)−1= (0, g1)·(1, g2) = (1, g1g2), and (1, g2)·(0, g1)−1= (1, g2)·(0, g−1
1) = (1, g1g2), so
these two are the same. There are m−krotations and kflips in A, so we thus expect the flips to
contribute 2(m−k)kto A+Abut only (m−k)kto A−A.
Next, consider the rotations in A+Aand A−A. Begin with F+Fand F−F. Since all flips
have order 2, these are in fact the same set and thus always contribute equally to A+Aand to
A−A. Also note that if k6= 0 (we will treat the k= 0 case later), the identity 1 is contained in
F+Fand F−F.