# The deformation of an erupting magnetic flux rope in a confined solar flare

Ruisheng Zheng1,2,3,⋆, Yihan Liu1, Liang Zhang1, Yang Liu1, Changhui Rao4,5,6,
Qing Lin7,
Zhimao Du7, Libo Zhong4,5, Huadong Chen3, Yao Chen1,2

###### Abstract

Magnetic flux ropes (MFRs), sets of coherently twisted magnetic field lines,
are believed as core structures of various solar eruptions. Their evolution
plays an important role to understand the physical mechanisms of solar
eruptions, and can shed light on adverse space weather near the Earth.
However, the erupting MFRs are occasionally prevented by strong overlying
magnetic fields, and the MFR evolution during the descending phase in the
confined cases is lack of attention. Here, we present the deformation of an
erupting MFR accompanied by a confined double-peaked solar flare. The first
peak corresponded to the MFR eruption in a standard flare model, and the
second peak was closely associated with the flashings of an underlying sheared
arcade (SA), the reversal slipping motion of the L-shaped flare ribbon, the
falling of the MFR, and the shifting of top of filament threads. All results
suggest that the confined MFR eruption involved in two-step magnetic
reconnection presenting two distinct episodes of energy release in the flare
impulsive phase, and the latter magnetic reconnection between the confined MFR
and the underlying SA caused the deformation of MFR.

Keywords: Sun: activity — Sun: flares — Sun: magnetic reconnection

1 Institute of Space Sciences, Shandong University, Weihai 264209, China
2 Institute of Frontier and Interdisciplinary Science, Shandong University,
Qingdao 266237, China
3 CAS Key Laboratory of Solar Activity, National Astronomical Observatories,
Beijing 100012, China
4 Institute of Optics and Electronics, Chinese Academy of Sciences, Chengdu
610209, China
5 The Key Laboratory on Adaptive Optics, Chinese Academy of Sciences, Chengdu
610209, China
6 University of Chinese Academy of Sciences, Beijing 100049, China
7 Shanghai Astronomy Museum, Shanghai 201306, China

##  1 Introduction

Magnetic flux ropes (MFRs) are regarded as coherent sets of twisted magnetic
field lines that wrap around their central axis, and their existence is
evidenced by filaments/prominences, coronal cavities, and sigmoid structures
in different emission lines. It is generally believed that MFRs are core
structures of various explosive solar activities. The evolution of MFRs plays
an important role to understand the physical mechanisms of solar eruptions,
and can be one main driver for adverse space weather near the Earth [1, 2, 3,
4, 5]. After the destabilization due to kink instability [6], torus
instability [7, 8], breakout reconnection [9], or tether-cutting reconnection
[10], a rising MFR will stretch the overlying magnetic field lines, and lead
to magnetic reconnection in the current sheet below, and finally cause a solar
flare that releases free magnetic energy [11, 12].

In the 3-dimensional standard flare model [13, 14], the MFR is surrounded by a
quasi-separatrix layers (QSLs) where the magnetic field connectivity changes
significantly [15]. When magnetic field lines pass through the QSLs,
successive reconnection will occur and result in the obvious slipping motion
along the QSLs [16, 17]. Due to slipping magnetic reconnection, the footpoints
of QSLs can be indicated by the observed flare ribbons that always show a form
of double J-shape structure. Hence, the eruption of the MFR surrounded by QSLs
is always accompanied by the slipping motions of flare ribbons and the
formation of hook structures at both ends of flare ribbons [18, 19].

Besides of the successful eruptions, the rising MFRs can also be prevented to
escape and cause confined eruptions [20, 21, 22, 23, 24, 25, 26, 27]. As
disclosed by observations and simulations, the failure of MFR eruptions are
predominantly due to the confinement of strong overlying magnetic fields as a
cage [28, 29, 30, 31]. Consequently, for a confined eruption, the rising MFR
experiences a deceleration and possibly falls back to the solar surface after
its stop at a certain height [20, 32].

As that in successful eruption, the development of a confined MFR is part of
the lifetime of the MFR, and is also important for understanding the physical
mechanisms of solar eruptions. However, the developments of MFRs in confined
eruptions are lack of attention. In this Letter, we focus on the evolution of
an erupting MFR in a confined double-peaked flare, and find the deformation of
the confined MFR by the external magnetic reconnection with underlying sheared
arcades.

##  2 Observations

The confined eruption occurred in NOAA AR 12860 ($\sim$S29W23) on 2021 August
29, and was linked to a double-peaked flare of C7.4 class
(SOL2021-08-29T00:44). The main observations of the eruption are from the
Atmospheric Imaging Assembly (AIA; [33]) onboard the Solar Dynamics
Observatory (SDO; [34]), and the AIA images has a pixel size of 0′′6 and a
cadence of 12 s. We also used observations from the Extreme Ultraviolet Imager
(EUVI; [35])-A on the the Solar Terrestrial Relations Observatory(STEREO;
[36]). The EUVI images have a pixel size of 1.58$arcsec$, and their cadences
are 2.5 minutes for the 195 Å. The X-ray images from the X-Ray Telescope (XRT;
[37]) aboard the Hinode [38] are collected to determine the X-ray structures
in AR 12860, with a pixel size of $\sim$1$arcsec$. The emission properties of
the eruption is also investigated with the differential emission measure (DEM)
method that employs the xrt_dem_iterative2.pro in SolarSoftWare package [39,
23]. In DEM method, the EM maps at different temperature ranges are obtained
by a set of AIA images in six channels (i.e., 94, 131, 171, 193, 211, and 335
Å).

The magnetic field evolution of AR 12860 is examined by full-disk magnetograms
and intensity images from the Helioseismic and Magnetic Imager (HMI; [40])
also onboard SDO, with a cadence of 45 s and pixel size of 0′′6\. The
filaments in AR 12860 were well captured by the H$\alpha$ filtergrams from the
Solar Magnetic Activity Research Telescope (SMART; [41]) at Hida observatory,
with a pixel size of $\sim$1$arcsec$. Moreover, the details of the sunspots
and filaments are complemented by the high-resolution images in H$\alpha$ and
TiO from the Educational Adaptive-optics Solar Telescope (EAST; [42]) that was
newly built in 2021 July at Shanghai Astronomy Museum, and their pixel
resolutions are 0′′12 for H$\alpha$ and TiO.

To check the configuration of the confined eruption, the nonlinear force-free
field (NLFFF) modeling [43, 44] is also utilized to construct the coronal
structures by setting the HMI photospheric vector magnetogram as the bottom
boundary. The NLFFF extrapolation is performed in a box of
116$\times$116$\times$117 uniformly distributed grid points with $\Delta
x=\Delta y=\Delta z=2arcsec$, and then we calculate the squashing factor $Q$
[15, 45] of the extrapolated magnetic field [46]. The radio dynamic spectra
associated with the eruption are obtained from the network of Compound
Astronomical Low frequency Low cost Instrument for Spectroscopy and
Transportable Observatory (CALLISTO; https://www.e-callisto.org/).

##  3 Results

###  3.1 Eruption and Deformation of MFR

Figure 1 shows the overview of AR 12860 and its magnetic field evolution
during the eruption. In the EAST TiO image (panel (a)), it is clear that AR
12860 is divided into two groups of sunspots, the preceding one (the orange
box) and the following one (the pink box). The high-resolution image of EAST
H$\alpha$ (panel (b)) clearly displays that a slender sigmoidal filament
suspended over the AR (green arrows), and some shorter filament threads (the
orange arrow) lay beneath the western portion of the sigmoidal filament. The
HMI intensity images and magnetograms (panels (c)-(f)) confirm that the
preceding group is dominated by negative polarities, and the following group
is comprised of predominant positive polarities and parasitic negative
polarities. Note that one parasitic sunspot (NS1, the blue arrow) emerged in
the west of following group and interacted with surrounding positive
polarities, which distorted the polarity inverse line in the following group.
In addition, NS1 also moved westwards as well as another negative sunspot
(NS2, the red arrow) in the preceding group. Along the shifting direction (the
green line in panel (e)), the time-distance plot (panel (g)) clearly shows the
sunspot movements with a speed of $\sim$100 m s-1 for NS1-NS2. The magnetic
flux variations (from 12:00 on August 28 to 06:00 UT on August 29) in the
source region of the eruption (red boxes in panels (d) and (f))) is shown in
panel (h). The unsigned negative flux kept increasing with a net increase of
$\sim 6\times 10^{20}Mx$, and the positive flux continuously decreased with a
net decrease of $\sim 5\times 10^{20}Mx$. It likely indicates an emergence and
cancellation of magnetic flux that continued through the period of the
eruption (from 00:20-00:50 UT, the green-shaded area in panel (h)). The
magnetic flux emergence and cancellation is consistent with the intrusion of
NS1 and the fading of the positive polarities in the source region (panels (d)
and (f)).

During the continuous magnetic activities and sunspot displacement, an MFR
formed in the AR and subsequently erupted (Figure 2 and Animation 1). The MFR
exhibited a transient sigmoid in high-temperature channels, similar to the
case in Kharayat et al. (2021), and its asymmetry was indicated by that the
western portion was much brighter than the eastern portion (panels (a)-(b)).
In the low corona and the chromosphere (panels (c)-(d)), the western portion
of the MFR was replaced by a twisted structure and a filament (green arrows),
and the eastern portion was hardly distinguished. At $\sim$00:25 UT on August
29, the asymmetric MFR erupted and resulted in a C7.4 flare. Unfortunately,
the eruption was seriously restricted by groups of overlying AR loops (panel
(c)). The flare induced a J-shaped ribbon and an L-shaped ribbon (the curves
in panel (g)). The L-shaped ribbon exhibited an obvious clockwise slipping
motion (the curved dashed arrow), and a hook structure (H1) appeared in the
extending end (the cyan arrow in the panel (h)). According to the
3-dimensional flare model [13, 14], the J- and L-shaped ribbon represented
footpoints of QSLs, and the appearance of the unusual L shape was possibly due
to the asymmetric eruption. The slipping motion of the L-shaped ribbon
resulted from magnetic reconnection successively occurring below the erupting
MFR.

Interestingly, the L-shaped ribbon sequentially experienced a reversal
slipping motion in a few minutes (Figure 3 and Animation 2). The anti-
clockwise slippage led a new hook structure (H2; the cyan arrow) in the east
of the L-shaped ribbon, accompanied by the darkening of the H1. Meanwhile, an
underlying sheared arcade (SA) appeared in the western part of the AR (blue
arrows). Intriguingly, during anti-clockwise slippage, some filament threads
(the green arrow) over the SA slowly moved southwards, and the SA became much
bright and showed some flashings (the pink arrows). Moreover, it is evident
that some plasmoids were released from the SA in AIA 193 Å (the red arrow in
panel (f)), similar with hot blobs reported in [26]. It likely indicates
magnetic reconnections between the filament threads and SA. Along the selected
path through two hook structures (S1, the green curve in panel (a)), the time-
distance plots show the bi-directional slipping motions in AIA 304 and 171 Å
(panels (g)-(h)). The clockwise slippage was prominent with a speed of
$\sim$80 km s-1 (the green dotted lines). The reversed slippage was weaker and
slower with a speed of $\sim$10 km s-1 (the green dashed lines). Note that the
reversed slippage closely followed with the movement of filament threads (the
white dashed line in panel (g)), which indicates a close relationship between
the reversed slippage and the movement of filament threads. Along the eruption
direction (S2, the red line in panel (c)), the time-distance plot shows the
evolution of the MFR (panel (i)). It is clear that the MFR first propagated a
distance of $\sim$50 Mm at a speed of $\sim$150 km s-1 (the white dotted line)
and began to fall down at $\sim$00:30 UT with a speed of $\sim$30 km s-1 (the
white dashed line). The falling of the MFR is consistent with the shifting of
filament threads in AIA 304 Å.

###  3.2 Two episodes of Magnetic Reconnection

The eruption is also shown in the limb view from EUVI-A and in composite
images in AIA 171 (red), 193 (green), and 94 (blue) Å (Figure 4 and Animation
3). In the limb view (panels (a)-(b)), the enveloping field lines in AR 12860
were nearly intact through the eruption, which possibly indicates a very
limited expansion of the enveloping loops. In the composite images (panels
(c)-(f), the high temperature component in 94 Å (blue) shows clearly the
evolution of the MFR from a transient sigmoid to a diffuse structure (red
arrows) that was restricted in the limited space experiencing a much less
expansion. Interestingly, the flashings (the blue arrow) on a sheared arcade
(SA) was surrounded by the diffuse high-temperature structure of the MFR,
which likely indicates the occurrence of the interaction between the expanding
MFR and with the underlying SA in the limited cage. After the interaction, the
newly-formed coronal loops (white arrows) became visible after the cooling of
the diffuse high-temperature structure ($\sim$ 01:20 UT).

The evolution of the flare ribbons is shown in Figure 5. The J- and L-shaped
flare ribbons were obvious in AIA 1600 Å (panel (a)), and their contours are
superposed on magnetic polarities of HMI magnetograms (panels (c) and (e)).
For the eruption region (the dashed box in panel (a)), the intensity curves in
AIA 1600, 304, and 94 Å (panel (b)) clearly reveal two peaks at $\sim$00:30
and $\sim$00:50 UT. It is consistent with the double peaks (dotted lines) of
the GOES 1-8 Å soft X-ray (SXR) flux and its derivative curve that indicate
the Neupert effect of the non-thermal and thermal flare emissions [47, 48].
Hence, the eruption indeed involved two episodes of magnetic reconnection. The
reconnected flux and the reconnection rate for the positive and negative
magnetic flux, estimated by the amount of magnetic flux swept by the flare
ribbons in the AIA 1600 Å, are shown with blue and red curves in panels (d)
and (f). The maximums of the reconnected flux through two episodes of
reconnection are separately $\sim 2\times 10^{20}Mx$ and $\sim 4\times
10^{19}Mx$. Around the peak time of two episodes of magnetic reconnection, two
Type III radio bursts (arrows in panel (g)) were recorded in the solar radio
dynamic spectra from CALLISTO.

Figure 6 shows the eruption in EM maps in different temperature ranges. At two
peak time of the double-peaked flare, the strongest emissions separately
reveal the post-flare loops (yellow arrows) and the SA (pink arrows) with
flashings. The post-flare loops are predominant in the high-temperature ranges
(left panels). The SA with flashings are dominated in the low-temperature
range (right panels). Note that the loops (white arrows in panels (f)) in
10-25 MK are consistent with the newly-formed loops in Figure 4(f), which
possibly indicates that they were the productions of magnetic reconnection
related to flashings. Hence, the eruption indeed involved two episodes of
magnetic reconnection.

##  4 Conclusions and Discussion

Combining with high-quality observations from EAST, HMI, and AIA, we report a
confined eruption of an asymmetry MFR. The eruption was closely related to the
continuous magnetic field activities (magnetic flux emergence and
cancellation) and the sunspot displacement (Figure 1). The confined eruption
was also asymmetry, which led to an L-shaped flare ribbon with a hook in the
predominant eruption direction and a normal J-shaped flare ribbon in the
opposite direction (Figure 2). Intriguingly, the L-shaped flare ribbon showed
a reversal slipping motion, and was accompanied by the slow shifting of
filament threads and the flashings of an underlying SA, which likely indicates
another magnetic reconnection (Figure 3). Two episodes of energy release
process for the confined eruption were indicated by AIA composite images and
EM maps, and were also evidenced by the double peaks in the EUV, SXR,
reconnected flux curves, and two type III radio bursts (Figure 4-6).

For the confined eruptions, it is believed that the upward force for rising
instable MFRs will finally be balanced by the strong confinement of overlying
field lines [20, 6, 29, 25, 27]. If an twisted MFR meets the overlying field
lines, the external magnetic reconnection possibly take places, and thus the
twist of MFR can be transferred to a larger coronal magnetic system [49, 50].
On the other hand, the confined MFR will stop at a certain height or fall back
to the solar surface, when there is no interaction with the overlying field
lines. Moreover, if one filament exists inside, the dark filament material can
be seen in the low-temperature wavelength during the falling of the MFR [20].
Due to the strong confinement, the confined MFR fell down (Figure 3(i)) and
possibly met with an underlying SA, indicated by the slow shifting of the low-
temperature counterpart (filament threads), the flashings, and the plasmoids
(Figure 3(d)-(f)). Moreover, along the L-shaped flare ribbon, the rising of
the erupting MFR caused the clockwise slipping motion, while the reversal
slipping motion occurred during the falling of the confined MFR (Figure
3(g)-(i)), which also infers the occurrence of magnetic reconnection. Hence,
in this Letter, the external magnetic reconnection occurred between the
confined MFR and the underlying SA.

Interestingly, the flare ribbons showed both the clockwise and anti-clockwise
slippage. The speed of the clockwise slippage is $\sim$80 km s-1, in the range
from several tens to $\sim$100 km s-1 for the typical slipping motion in the
3-dimensional flare model [18, 19], whereas the speed of the anti-clockwise
slippage is only $\sim$10 km s-1 (Figure 3(g)-(f)). The significant difference
on the slippage speed is likely resulted from two kinds of magnetic
reconnection, consistent with the different reconnection flux and rates
(Figure 5 (d) and (f)). The fast clockwise slippage is likely induced by the
successive reconnection between two groups of arcades overlying the rising MFR
in the impulsive eruption stage, and the slower anti-clockwise slippage
possibly indicates an intermittent reconnection between the confined falling
MFR and the underlying SA. Hence, it is possible that the reversal slippage
represents the shrinking of the footprints of the confined MFR following the
latter external reconnection.

The double-peaked feature of the flare is the indicator of two-step magnetic
reconnection involved with the confined eruption. The first peak corresponded
to the internal magnetic reconnection between the stretching anti-parallel
field lines in a standard flare model, and the second peak was closely
associated with the external magnetic reconnection between the confined MFR
and the underlying SA. In the derivative of SXR flux (Figure 4(g)), the first
peak is much stronger than the second peak, which possibly means that the non-
thermal emissions in the internal magnetic reconnection were much intense than
that in the external magnetic reconnection.

It is reasonable that two-step magnetic reconnection releases energy in two
episodes in the form of double peaks of the SXR flare. However, it is not
often that the eruption involved with two-step magnetic reconnection must be
accompanied by a double-peaked SXR flare. In some cases with M- and X-class
flares [51, 52, 53, 54], the SXR flux only exhibited a short bump or nothing
in the impulsive rising phase that lasted for a few minutes. The impulsive
phase of the double-peaked flare in this Letter had a period of $\sim$24
minutes in GOES SXR and $\sim$34 minutes in AIA 94 Å (yellow and cyan shades
in Figure 5(b)), which is similar to that in [55]. Hence, we suggest that, for
the flare involving with two-step magnetic reconnection, the longer the
duration of the impulsive phase is, the clearer the profile of the double
peaks becomes.

In addition, we check the configuration of the source region before the
eruption (00:00 UT) with the NLFFF model in Figure 7. The extrapolated fields
are superposed on HMI vector magnetogram in the side view (left) and on
contours of $Q$ value at the photospheric surface in the top view (right). The
orange group of filed lines represents the MFR, and two groups of overlying
loops are indicated by the blue and red field lines. The high $Q$ value
beneath the MFR center indicates the possible sites of the internal
reconnection, though the value is much less than that in earlier reports [56,
57]. The forward and reverse slipping motions are also superimposed on the
counters of $Q$ value (yellow and cyan dashed arrows).

Based on the above results and discussions, we propose a possible scenario for
the confined eruption in the schematic drawings in Figure 8. At the eruption
stage (panels (a)-(b)), the rising MFR (red lines) stretched two groups of
overlying loops (blue and green lines), and initiated the internal magnetic
reconnection (the star symbol) that occurred beneath the rising MFR between
two groups of overlying loops, following the 3-dimensional flare model. As a
result, magnetic reconnection induced the lower post-flare loops and the
higher and longer loops wrapping the rising MFR (blue-green lines), and
J-shaped and L-shaped flare ribbons (purple curves). Due to the asymmetric
eruption, the L-shaped ribbon appeared and exhibited the slipping motion (the
clockwise arrow). The eruption was confined by the higher overlying strong
cage (orange). At the post-confinement stage (panels (c)-(d)), the SA (the
black line) appeared beneath the confined MFR, and the external magnetic
reconnection (the star symbol) occurred between the MFR and the underlying SA.
The external reconnection was accompanied by the reversal slipping motion (the
anti-clockwise arrow) of the L-shaped ribbon, and the MFR finally deformed as
two newly-formed loops (pink-black lines).

Most of studies for MFRs focus on the evolution of the formation and eruption
[20, 6, 29, 25, 27], and the MFR evolution during the descending phase in the
confined case, as the primary aim of this Letter, is lack of attention. In
summary, the confined eruption of the MFR involved with two-step magnetic
reconnection relating to two distinct episodes of energy release in the flare
impulsive phase, and the latter external reconnection led to the deformation
of the confined MFR. We propose that the deformation of confined MFRs can
complement the understanding of the entire evolution of MFRs and the
initiation of solar eruptions. Further observations and simulations are
desirable in the future.

Acknowledgements

The authors thank the anonymous referee for constructive comments and Dr.
Zhenghua Huang for the helpful discussion. We gratefully acknowledge the use
of data from Shanghai Astronomy Museum’ s Educational Adaptive-optics Solar
Telescope (EAST) built by Institute of Optics and Electronics, Chinese Academy
of Sciences. The EAST operation is supported by the Shanghai Science &
Technology Museum. SDO is a mission of NASA’s Living With a Star Program. The
authors thank the teams of SDO, SMART, and EAST for providing the data. This
work is supported by grants NSFC 11790303, 11727805, and 12073016, and the
open topic of the Key Laboratory of Solar Activities of Chinese Academy
Sciences (KLSA202108).

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Figure 1: The overview of AR 12860 and its magnetic field evolution. ((a)-(f))
The orange and the pink dashed boxes separately outline the preceding sunspots
and following sunspots, and the blue and red arrows points out the moving
negative sunspots, NS1 and NS2. The green and orange arrows show the sigmoidal
filament and the underlying threads. (g) Time-distance plot along the green
line in panel (e), showing the movement of NS1 and NS2. The dotted lines are
used to derive the attached speeds. (h) The negative (red) and positive (blue)
magnetic flux evolution (between 12:00 on August 28 to 06:00 UT on August 29)
in the AR center (red boxes in panel (d) and (f)). The green-shaded area
represents the duration of the eruption.  Figure 2: The asymmetric MFR before
the eruption (upper panels) and flare ribbons after the eruption (bottom
panels). The red box points out the field of view of other panels. The curved
lines represent the J-shaped and inverse-L-shaped flare ribbons, and the
curved arrow shows the clockwise slipping motion. The green and cyan arrows
separately indicate the western part of the MFR and the first hook (L1). (An
animation is available online, and the animated sequence runs from 23:35 on
August 28 to 00:35 UT on August 29.)  Figure 3: ((a)-(f)) The deformation of
the confined MFR in AIA 304, 171, 193, and 131 Å. The blue arrows indicate the
underlying SA, and the pink arrows show the flashings on the SA. The green and
cyan arrows point out the shifting filament threads and the second hook (H2),
respectively. The red arrow points out the plasmoids. The curved arrow
represents the anti-clockwise slipping motion. ((g)-(i)) Time-distance plots
along the curve (S1) in panel (a) and the line (S2) in panel (c), uncovering
the bi-directional slipping motions and the MFR eruption. The dotted and
dashed lines are used to derive the attached speeds. (An animation of AIA
images is available online, and the animated sequence runs from 00:35 to 01:00
UT on August 29.)  Figure 4: The eruption in two perspectives of EUVI-A 195 Å
((a)-(b)) and the composite images of AIA 171 (red), 193 (green), and 94
(blue) Å ((c)-(f)). The deformation of the confined MFR in AIA 304, 171, 94 Å.
The red and blue arrows separately indicate the MFR and the SA. The white
arrows show the newly-formed loops after the eruption. (An animation with AIA
and EUVI-A images is available online, and the animated sequence runs from
00:00 to 01:00 UT on August 29.)  Figure 5: The flare ribbons in AIA 1600 Å
and their contours over HMI magnetograms (left panels). (b) The intensity
curves in AIA 1600, 304, and 94 Å for the eruption region (the red box in
panel (a)) and the GOES SXR flux and its derivative for the flare. The
vertical lines mark the two peak times of the flare. The yellow and cyan
shades separately cover the flare impulsive phase in GOES SXR and AIA 94 Å.
((d) and (f)) The curves of the reconnected flux and the reconnection rate for
the positive (blue) and negative (red) magnetic flux. (g) The radio dynamic
spectra from CALLISTO, showing the type III radio bursts (white arrows).
Figure 6: Two-step magnetic reconnection in EM maps at temperature range of
0.3-1 MK, 1-10 MK, and 10-25 MK. The yellow and pink arrows separately
indicate the post-flare loops and the flashings. The white arrows show the
newly-formed high-temperature loops.

Figure 7: The NLFFF extrapolated MFR (orange) and the overlying arcades (blue
and red) before the eruption (00:00 UT) superposed on the HMI vector
magnetogram in the side view (left) and on the contours of $Q$ value at
photospheric surface in the top view (right). The yellow and cyan dashed
arrows show two kinds of slipping motions.

Figure 8: Schematic drawings for the eruption and deformation of the MFR (red)
below the confining cage (orange) based on the contours of positive and
negative polarities from the HMI magnetogram at 00:00 UT on August 29. The
green and blue lines represent the filed lines stretched by the rising MFR,
and the green-blue lines show the post-flare loops and the newly-formed lines
wrapping the rising MFR due to the inner magnetic reconnection. The underlying
SA is indicated by the short black line, and the pink-black lines show the
newly-formed loops resulted from the external magnetic reconnection. The flare
ribbons are outlined by the purple curves, and the two-step magnetic
reconnection is indicated by star symbols.


# On the Diameter of Undirected Cayley Graphs of Finite Abelian Groups

Béla Bajnok111Department of Mathematics, Gettysburg College, U.S.A. Email:
<EMAIL_ADDRESS>W. Kyle Beatty222Department of Applied Mathematics and
Statistics, Johns Hopkins University, U.S.A. Email<EMAIL_ADDRESS>
(April 1, 2024)

###### Abstract

Let $s$ be a positive integer. Our goal is to find all finite abelian groups
$G$ that contain a $2$-subset $A$ for which the undirected Cayley graph
$\Gamma(G,A)$ has diameter at most $s$. We provide a complete answer when $G$
is cyclic, and a conjecture and some partial answers when $G$ is noncyclic.

2020 AMS Mathematics Subject Classification: Primary: 11B13; Secondary: 05B10,
05C35, 11P70, 11B75, 20K01.

Key words: Abelian group, Cayley graph, diameter, sumset, signed sumset.

##  1 Introduction

Throughout this paper we let $G$ be an additively written finite abelian
group. When $G$ is cyclic and of order $n$, we identify it with
$\mathbb{Z}_{n}=\mathbb{Z}/n\mathbb{Z}$; we consider $0,1,\dots,n-1$
interchangeably as integers and as elements of $\mathbb{Z}_{n}$. Below we use
standard terminology and notation of additive combinatorics (see [1]).

Given a finite abelian group $G$ and a subset $A\subseteq G$, the undirected
Cayley graph $\Gamma(G,A)$ has its vertex set formed by the elements of $G$,
and has an edge between two vertices whenever their difference is an element
of $A$, that is, $g_{1}\in G$ and $g_{2}\in G$ are connected when
$g_{2}=g_{1}+a$ or $g_{1}=g_{2}+a$ for some $a\in A$. If $A$ is a generating
set of $G$, then $\Gamma(G,A)$ is connected; in particular, if $A$ is an
$s$-spanning set of $G$, meaning that every element of $G$ can be written as
the signed sum of at most $s$ (not-necessarily distinct) elements of $A$, then
$\Gamma(G,A)$ has diameter at most $s$.

Suppose now that $A$ is an $m$-subset of $G$. It is easy to compute that there
are at most $\sum_{i\geq 0}{m\choose i}{h-1\choose i-1}2^{i}$ vertices of
$\Gamma(G,A)$ that are of distance $h$ away from any given vertex, and thus if
$A$ is an $s$-spanning set in $G$, then the order of $G$ can be at most

$\sum_{h=0}^{s}\sum_{i\geq 0}{m\choose i}{h-1\choose i-1}2^{i}=\sum_{i\geq
0}{m\choose i}{s\choose i}2^{i}.$

Therefore, we have the following bound.

###### Proposition 1.1

For positive integers $m$ and $s$, define

$\displaystyle a(m,s)=\sum_{i\geq 0}{m\choose i}{s\choose i}2^{i}.$ (1)

If an abelian group $G$ has an $m$-subset $A$ for which the diameter of
$\Gamma(G,A)$ is $s$, then the order of $G$ is at most $a(m,s)$.

We note, in passing, that the values of $a(m,s)$ are known as Delannoy numbers
(see [3]), and that they satisfy the recursion

$a(m,s)=a(m,s-1)+a(m-1,s)+a(m-1,s-1).$

It is a natural question to ask for the largest possible order of a group $G$
with an $m$-subset $A$ for which the diameter of ${\Gamma}(G,A)$ is at most
$s$. Before we discuss our results on this question, let us review the
analogous question in the directed case.

Given a finite abelian group $G$ and a subset $A\subseteq G$, the directed
Cayley graph $\vec{\Gamma}(G,A)$ has its vertex set formed by the elements of
$G$, and has a directed edge from $g_{1}\in G$ to $g_{2}\in G$ when
$g_{2}=g_{1}+a$ for some $a\in A$. We say that $A$ is an $s$-basis of $G$ if
every element of $G$ can be written as the sum of at most $s$ (not-necessarily
distinct) elements of $A$; in this case, the diameter of $\vec{\Gamma}(G,A)$
is at most $s$.

For an $m$-subset $A$ of $G$, we see that there are at most ${m+h-1\choose h}$
vertices of $\vec{\Gamma}(G,A)$ that are of distance $h$ away from any given
vertex, so if $A$ is an $s$-basis of $G$, then we must have

$|G|\leq\sum_{h=0}^{s}{m+h-1\choose h}={m+s\choose s}.$

When equality holds, we say that $A$ is a perfect $s$-basis of $G$: in this
case, every element of $G$ arises as the sum of at most $s$ (not-necessarily
distinct) elements of $A$ in a unique way. Trivially, the set of nonzero
elements is a perfect $1$-basis in any group $G$, and the $1$-element set
consisting of a generator of $G$ is a perfect $s$-basis in the cyclic group of
order $s+1$. It turns out that there are no others:

######  Theorem 1.2 (Bajnok, Berson, and Just; cf. [2])

If a subset $A$ of a finite abelian group $G$ is a perfect $s$-basis, then
$s=1$ and $A=G\setminus\\{0\\}$, or $G\cong\mathbb{Z}_{s+1}$ and $|A|=1$.

Given the scarcity of perfect bases, it is natural to ask for the largest
groups with a given diameter and size of generating set. The paper [9] asks
the following questions:

  * •

What is the largest possible order of an abelian group $G$ with an $m$-subset
$A$ for which the diameter of $\vec{\Gamma}(G,A)$ is at most $s$?

  * •

What is the largest possible order of a cyclic group $G$ with an $m$-subset
$A$ for which the diameter of $\vec{\Gamma}(G,A)$ is at most $s$?

After a series of papers [10, 5, 8, 9, 4], these two questions have been
completely solved for the case of $m=2$. We have the following two results.

######  Theorem 1.3 (Morillo, Fiol, and Fàbrega; cf. [10])

Let $s$ be a positive integer. The largest possible order of a cyclic group
$G$ with a $2$-subset for which the diameter of $\vec{\Gamma}(G,A)$ is at most
$s$ is given by $\left\lfloor(s^{2}+4s+3)/3\right\rfloor$.

######  Theorem 1.4 (Fiol; cf. [4])

Let $s$ be any positive integer. The largest possible order of an abelian
group $G$ with a $2$-subset for which the diameter of $\vec{\Gamma}(G,A)$ is
at most $s$ is given by $\left\lfloor(s^{2}+4s+4)/3\right\rfloor$.

Observe that the quantities in Theorems 1.3 and 1.4 agree, unless $s\equiv 1$
mod 3; for that case one can verify that the group
$G=\mathbb{Z}_{k}\times\mathbb{Z}_{3k}$ has the $s$-basis
$\\{(0,1),(1,3k-1)\\}$, where $k=(s+2)/3$. However, the question specifically
for noncyclic groups is still open when $s\not\equiv 1$ mod 3.

Let us now return to the undirected case. When equality occurs in Proposition
1.1, then $A$ is called a perfect $s$-spanning set in $G$. Perfect spanning
sets can be easily characterized for $m=1$ and $s=1$. Note that $a(1,s)=2s+1$
and, clearly, for any $a\in\mathbb{Z}_{2s+1}$ with $\gcd(a,2s+1)=1$, the set
$\\{a\\}$ is a perfect $s$-spanning set in $\mathbb{Z}_{2s+1}$. Also,
$a(m,1)=2m+1$, and for any group $G$ of order $2m+1$, every asymmetric subset
of size $m$ (that is, a set containing exactly one of any nonzero element or
its inverse) is a perfect $1$-spanning set in $G$.

From (1) we get $a(2,s)=2s^{2}+2s+1$, and below we establish the following
result.

###### Theorem 1.5

Let $s$ be a positive integer. The largest possible order of a cyclic group
$G$ with a $2$-subset for which the diameter of ${\Gamma}(G,A)$ is at most $s$
is given by $2s^{2}+2s+1$.

For noncyclic groups the situation seems a bit more nuanced. We have the
following conjecture.

###### Conjecture 1.6

Let $s\geq 3$ be an integer.

  * •

If $2s+1$ is prime, then the largest noncyclic group with an $s$-spanning set
of size $2$ has order $2s^{2}$.

  * •

If $2s+1$ is composite and $p$ is its smallest prime divisor, then the largest
noncyclic group with an $s$-spanning set of size $2$ has order
$2s^{2}+2s-(p^{2}-1)/2$.

In particular, for any noncyclic group $G$ with an $s$-spanning set of size
$2$ we have $|G|\leq 2s^{2}+2s-4.$

(We add that the largest noncyclic group with a $2$-spanning set of size $2$
is $\mathbb{Z}_{3}\times\mathbb{Z}_{3}$.)

Below we prove that the quantities in Conjecture 1.6 are lower bounds.

###### Theorem 1.7

Let $s$ be a positive integer.

  * •

The largest noncyclic group with an $s$-spanning set of size $2$ has order at
least $2s^{2}$.

  * •

If $2s+1$ is composite and $p$ is any prime divisor of $2s+1$, then the
largest noncyclic group with an $s$-spanning set of size $2$ has order at
least $2s^{2}+2s-(p^{2}-1)/2$.

Establishing upper bounds appears to be more difficult, and we only have the
partial results that we summarize here. We first note that a noncyclic group
that has a spanning set of size $2$ must have rank $2$ and thus is isomorphic
to $\mathbb{Z}_{c}\times\mathbb{Z}_{kc}$ for some integers $c\geq 2$ and
$k\geq 1$.

We say that the group $\mathbb{Z}_{c}\times\mathbb{Z}_{kc}$ is $s$-regular for
some positive integer $s$, if it possesses an $s$-spanning set of the form
$\\{(1,u),(1,v)\\}$ for some $u,v\in\mathbb{Z}_{kc}$. While we have no precise
count on the number of $s$-regular groups, it appears that most groups of rank
$2$ that have an $s$-spanning set of size $2$ are $s$-regular. For example,
for $s=10$, of the $105$ groups of rank $2$ that have an $s$-spanning set of
size $2$, $103$ are $s$-regular. (The two exceptions are
$\mathbb{Z}_{8}\times\mathbb{Z}_{16}$ and
$\mathbb{Z}_{10}\times\mathbb{Z}_{20}$.) In Theorem 3.1 below we prove for
every odd prime $c$ and positive integer $k$ that if
$\mathbb{Z}_{c}\times\mathbb{Z}_{kc}$ has an $s$-spanning set at all, then it
is $s$-regular.

Our results for upper bounds can be summarized as follows.

###### Theorem 1.8

Let $s$ be a positive integer, and let $G=\mathbb{Z}_{c}\times\mathbb{Z}_{kc}$
be an $s$-regular group for some integers $c\geq 2$ and $k\geq 1$.

  * •

If $c=3$ and $s+1$ is divisible by $3$, then $|G|\leq 2s^{2}+1$.

  * •

If $c$ is odd and $s$ is divisible by $c$, then $|G|\leq 2s^{2}+s$, and if
$2s+1$ is divisible by $c$, then $|G|\leq 2s^{2}+2s-(c^{2}-1)/2$.

  * •

In all other cases, $|G|\leq 2s^{2}$.

As a consequence of our results, we can prove that for $s\geq 3$, every
$s$-regular abelian group of rank 2 has order at most $2s^{2}+2s-4$, with
equality if, and only if, it is isomorphic to
$\mathbb{Z}_{3}\times\mathbb{Z}_{(2s^{2}+2s-4)/3}$ for some $s\equiv 1$ mod
$3$ (See Theorem 3.3 below).

For $c=2$ we have exact results, even without the assumption that the group is
$s$-regular:

###### Theorem 1.9

Let $k$ and $s$ be positive integers. The largest possible order of a group of
the form $\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$ that has an $s$-spanning set of
size $2$ is $2s^{2}$ when $s$ is even and $2s^{2}-2$ when $s$ is odd.

In Section 4 below we provide some further results on groups of the form
$\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$.

For $m\geq 3$, the largest order of abelian groups with an $m$-subset for
which the diameter of their Cayley graphs is at most $s$ is generally unknown,
though some small cases for cyclic groups were displayed by Haanpää in [7], by
Hsu and Jia in [8], and by Graham and Sloane in [6].

Of particular interest are perfect spanning sets. An $m$-subset $A$ of $G$ is
called a perfect $s$-spanning set, if every element of $G$ arises as a signed
sum of at most $s$ terms in a unique way; in this case we have $|G|=a(m,s)$.
Above we mentioned perfect spanning sets for $m=1$, for $s=1$, and for $m=2$;
we believe that there are no others.

###### Conjecture 1.10

If an $m$-subset $A$ of a finite abelian group $G$ is a perfect $s$-spanning
set, then one of the following holds:

  * •

$m=1$, $G\cong\mathbb{Z}_{2s+1}$, and $A=\\{a\\}$ for some
$a\in\mathbb{Z}_{2s+1}$ with $\gcd(a,2s+1)=1$;

  * •

$s=1$, and $G$ is the disjoint union of $\\{0\\}$, $A$, and $-A$; or

  * •

$m=2$, $G\cong\mathbb{Z}_{2s^{2}+2s+1}$, and $A=d\cdot\\{s,s+1\\}$ for some
$d\in\mathbb{Z}_{2s+1}$ with $\gcd(d,2s^{2}+2s+1)=1$.

In Sections 2 and 3 below, we present our results on lower bounds and upper
bounds, respectively. In Section 4 we treat the case of the groups
$\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$ in more detail.

##  2 Constructions and lower bounds

In this section we provide some explicit constructions for $s$-spanning sets
of size $2$. We denote the $s$-span of a 2-subset $A=\\{a_{1},a_{2}\\}$ of $G$
by

$\langle
A\rangle_{s}=\\{\lambda_{1}a_{1}+\lambda_{2}a_{2}\mid\lambda_{1},\lambda_{2}\in\mathbb{Z},|\lambda_{1}|+|\lambda_{2}|\leq
s\\}.$

It will be helpful to have the following facts available.

###### Lemma 2.1

For a given positive integer $s$, let

$I(s)=\\{(\lambda_{1},\lambda_{2})\mid\lambda_{1},\lambda_{2}\in\mathbb{Z},|\lambda_{1}|+|\lambda_{2}|\leq
s\\}.$

  1. 1.

For a given integer $h$ with $|h|\leq s$, the number of ordered pairs
$(\lambda_{1},\lambda_{2})\in I(s)$ with $\lambda_{1}+\lambda_{2}=h$ equals
$s+1$ when $s$ and $h$ have the same parity, and $s$ otherwise.

  2. 2.

There are $(s+1)^{2}$ ordered pairs $(\lambda_{1},\lambda_{2})\in I(s)$ where
$\lambda_{1}+\lambda_{2}$ and $s$ have the same parity, and $s^{2}$ where
$\lambda_{1}+\lambda_{2}$ and $s$ have opposite parity.

  3. 3.

There are $2s^{2}+2s+1$ elements in $I(s)$.

Proof. By symmetry, we may assume that $h\geq 0$. Observe that the ‘leftmost’
integer point $(\lambda_{1},\lambda_{2})$ on the line $x+y=h$ within
$|\lambda_{1}|+|\lambda_{2}|\leq s$ is

$\left(-\left\lfloor(s-h)/2\right\rfloor,\;h+\left\lfloor(s-h)/2\right\rfloor\right),$

and the ‘rightmost’ such point is

$\left(\left\lfloor(s+h)/2\right\rfloor,\;h-\left\lfloor(s+h)/2\right\rfloor\right),$

a set of $s+1$ points when $s$ and $h$ have the same parity, and $s$ points
otherwise. This establishes the first claim, from which our other two claims
follow easily. $\Box$

###### Proposition 2.2

For all positive integers $s$ and $n$ with

$2\leq n\leq 2s^{2}+2s+1,$

the set $\\{s,s+1\\}$ is an $s$-spanning set in $\mathbb{Z}_{n}$.

Proof. The fact that $n$ cannot be larger than $2s^{2}+2s+1$ follows from
Proposition 1.1. We verify that for every $n$ in the range given above, we
have $\langle\\{s,s+1\\}\rangle_{s}=\mathbb{Z}_{n}.$

Considering the elements of

$S=\\{\lambda_{1}s+\lambda_{2}(s+1)\mid\lambda_{1},\lambda_{2}\in\mathbb{Z},|\lambda_{1}|+|\lambda_{2}|\leq
s\\}$

in $\mathbb{Z}$ (rather than $\mathbb{Z}_{n}$), we see that they lie in the
interval $[-(s^{2}+s),(s^{2}+s)]$. Since the index set

$I(s)=\\{(\lambda_{1},\lambda_{2})\mid\lambda_{1},\lambda_{2}\in\mathbb{Z},|\lambda_{1}|+|\lambda_{2}|\leq
s\\}$

contains exactly $2s^{2}+2s+1$ elements (see Lemma 2.1), it suffices to prove
that no integer in $[-(s^{2}+s),(s^{2}+s)]$ can be written as an element of
$S$ in two different ways.

For that, suppose that

$\lambda_{1}s+\lambda_{2}(s+1)=\lambda_{1}^{\prime}s+\lambda_{2}^{\prime}(s+1)$

for some $(\lambda_{1},\lambda_{2})\in I(s)$ and
$(\lambda_{1}^{\prime},\lambda_{2}^{\prime})\in I(s)$; without loss of
generality, we can assume that $\lambda_{2}\geq\lambda_{2}^{\prime}$. Our
equation implies that $\lambda_{2}-\lambda_{2}^{\prime}$ is divisible by $s$
and is, therefore, equal to 0, $s$, or $2s$.

If $\lambda_{2}-\lambda_{2}^{\prime}=2s,$ then $\lambda_{2}=s$ and
$\lambda_{2}^{\prime}=-s$; since $|\lambda_{1}|+|\lambda_{2}|\leq s$ and
$|\lambda_{1}^{\prime}|+|\lambda_{2}^{\prime}|\leq s$, this can only happen if
$\lambda_{1}=\lambda_{1}^{\prime}=0$, but this case leads to a contradiction
with our equation.

Assume, next, that $\lambda_{2}-\lambda_{2}^{\prime}=s.$ Our equation then
yields $\lambda_{1}-\lambda_{1}^{\prime}=-s-1.$ In this case, we have
$\lambda_{1}\leq 0$, $\lambda_{1}^{\prime}\geq 0$, $\lambda_{2}\geq 0$, and
$\lambda_{2}^{\prime}\leq 0$, and we see that

$|\lambda_{1}|+|\lambda_{2}|=-\lambda_{1}+\lambda_{2}=(s+1-\lambda_{1}^{\prime})+(s+\lambda_{2}^{\prime})=2s+1-(|\lambda_{1}^{\prime}|+|\lambda_{2}^{\prime}|)\geq
s+1,$

but that is a contradiction.

This leaves us with the case that $\lambda_{2}=\lambda_{2}^{\prime}$, which
implies $\lambda_{1}=\lambda_{1}^{\prime}$, as claimed, and therefore the set
$\\{s,s+1\\}$ is $s$-spanning in $\mathbb{Z}_{n}$. $\Box$

###### Proposition 2.3

For every positive integer $s$, the set $\\{(0,1),(1,1)\\}$ is an $s$-spanning
set in $\mathbb{Z}_{s}\times\mathbb{Z}_{2s}$.

Proof. We need to prove that for any
$(a,b)\in\mathbb{Z}_{s}\times\mathbb{Z}_{2s}$, we have
$(a,b)\in\langle\\{(0,1),(1,1)\\}\rangle_{s}.$ Note that by symmetry we may
assume that $0\leq b\leq s$. We consider three subcases.

When $b\geq a$, we let $\lambda_{1}=b-a$ and $\lambda_{2}=a$. We then have

$\lambda_{1}(0,1)+\lambda_{2}(1,1)=(a,b)$

and

$|\lambda_{1}|+|\lambda_{2}|=\lambda_{1}+\lambda_{2}=b\leq s,$

so $(a,b)\in\langle\\{(0,1),(1,1)\\}\rangle_{s}.$

Next, when $2a-s\leq b<a$, we again let $\lambda_{1}=b-a$ and $\lambda_{2}=a$.
In this case, we have

$|\lambda_{1}|+|\lambda_{2}|=-\lambda_{1}+\lambda_{2}=2a-b\leq s,$

so again $(a,b)\in\langle\\{(0,1),(1,1)\\}\rangle_{s}.$

Finally, when $b<2a-s$, we let $\lambda_{1}=b-a+s$ and $\lambda_{2}=a-s$. Here
we have

$\lambda_{1}(0,1)+\lambda_{2}(1,1)=(a-s,b),$

where $a-s\equiv a$ mod $s$. Furthermore,

$|\lambda_{1}|+|\lambda_{2}|=\lambda_{1}-\lambda_{2}=(b-a+s)+(s-a)<s,$

so again $(a,b)\in\langle\\{(0,1),(1,1)\\}\rangle_{s}.$ This completes our
proof. $\Box$

###### Theorem 2.4

Let $c$ and $s$ be positive integers so that the remainder of $s$ mod $c$ is
at least $(c-1)/2$. Let $u=\lfloor s/c\rfloor$, $v=\lceil s/c\rceil$, and
$k=2uv$. Then $A=\left\\{(1,u),(1,v)\right\\}$ is an $s$-spanning set in
$G=\mathbb{Z}_{c}\times\mathbb{Z}_{ck}$.

Proof. Let $H=\langle(1,u)\rangle$ be the subgroup of $G$ generated by the
element $(1,u)$. Since $H$ has order $2cv$, it has $cu$ cosets in $G$. We
claim that these cosets are $\lambda_{2}\cdot(1,v)+H$, where
$\lambda_{2}=0,1,\ldots,cu-1$. In order to prove this, we need to show that
these $cu$ cosets are pairwise distinct.

Suppose, then, that $\mu_{1}(1,v)$ and $\mu_{2}(1,v)$ are in the same coset of
$H$ for some $0\leq\mu_{1},\mu_{2}\leq cu-1$. This implies that
$(\mu_{2}-\mu_{1})\cdot(1,v)$ is an element of $H$, so there exists an integer
$\mu$ for which

$(\mu_{2}-\mu_{1})\cdot(1,v)=\mu\cdot(1,u).$

From the first coordinates, this means that $\mu_{2}-\mu_{1}-\mu$ is divisible
by $c$, and from the second coordinates we see that $(\mu_{2}-\mu_{1})v-\mu
u$, which, since $v=u+1$, equals $(\mu_{2}-\mu_{1}-\mu)u+(\mu_{2}-\mu_{1}),$
is divisible by $ck$ and thus by $cu$. Combining these, we get that
$\mu_{2}-\mu_{1}$ must also be divisible by $cu$, but that is only possible if
$\mu_{1}=\mu_{2}$.

Let $g\in G$ be arbitrary. According to our claim, we have integers
$\lambda_{1}\in[-cv+1,cv]$ and $\lambda_{2}\in[0,cu-1]$ for which

$g=\lambda_{1}(1,u)+\lambda_{2}(1,v).$

By assumption, we have integers $q$ (quotient) and $r$ (remainder) with
$(c-1)/2\leq r\leq c-1$ for which $s=qc+r$. We then have

$\displaystyle|\lambda_{1}|+|\lambda_{2}|\leq
cv+(cu-1)=c(q+1)+(cq-1)=2cq+c-1=2s+(c-2r-1)\leq 2s.$ (2)

If $|\lambda_{1}|+|\lambda_{2}|\leq s$, then $g\in\langle A\rangle_{s}$, and
we are done.

Assume then that $|\lambda_{1}|+|\lambda_{2}|\geq s+1$. In this case, let
$\lambda_{1}^{\prime}=\lambda_{1}-\epsilon cv$ and
$\lambda_{2}^{\prime}=\lambda_{2}-cu$, where $\epsilon$ equals $1$ when
$\lambda_{1}\geq 0$, and $-1$ when $\lambda_{1}<0$. We can compute that

$\lambda_{1}^{\prime}(1,u)+\lambda_{2}^{\prime}(1,v)=(\lambda_{1}-\epsilon
cv)(1,u)+(\lambda_{2}-cu)(1,v)=\lambda_{1}(1,u)+\lambda_{2}(1,v)-(\epsilon
cv+cu,(\epsilon+1)cuv),$

and since $\epsilon+1$ equals $0$ or $2$, in $G$ we have

$\lambda_{1}^{\prime}(1,u)+\lambda_{2}^{\prime}(1,v)=\lambda_{1}(1,u)+\lambda_{2}(1,v)=g.$

Finally, we show that $|\lambda_{1}^{\prime}|+|\lambda_{2}^{\prime}|\leq s$.
Indeed, we have

$|\lambda_{1}^{\prime}|+|\lambda_{2}^{\prime}|=|\lambda_{1}-\epsilon
cv|+|\lambda_{2}-cu|=cv-|\lambda_{1}|+cu-|\lambda_{2}|.$

From (2) above we see that $cv+cu\leq 2s+1$, and by our current assumption we
have $|\lambda_{1}|+|\lambda_{2}|\geq s+1$. Therefore,
$|\lambda_{1}^{\prime}|+|\lambda_{2}^{\prime}|\leq s$, as claimed, and our
proof is now complete. $\Box$

###### Corollary 2.5

Let $s$ be a positive integer for which $2s+1$ is composite, and let $p$ be
any prime divisor of $2s+1$. Then the largest noncyclic group with an
$s$-spanning set of size $2$ has order at least $2s^{2}+2s-(p^{2}-1)/2.$

Proof. Let us write $2s+1=pt$ for some integer $t\geq 2$. Note that $p$ and
$t$ must be odd, and we have

$s=\frac{t-1}{2}p+\frac{p-1}{2},$

$\lfloor s/p\rfloor=(t-1)/2$, and $\lceil s/p\rceil=(t+1)/2$. Let
$G=\mathbb{Z}_{p}\times\mathbb{Z}_{pk}$ where $k=(t^{2}-1)/2$. Then $G$ has
order $2s^{2}+2s-(p^{2}-1)/2$ and, according to Theorem 2.4, it has an
$s$-spanning set of size $2$. $\Box$

##  3 $s$-regular groups and upper bounds

Recall that we say that the group $\mathbb{Z}_{c}\times\mathbb{Z}_{kc}$ is
$s$-regular for some positive integer $s$, if it possesses an $s$-spanning set
of the form $\\{(1,u),(1,v)\\}$ for some $u,v\in\mathbb{Z}_{kc}$. In this
section we provide some upper bounds for the largest possible order of an
$s$-regular group.

###### Theorem 3.1

Let $G=\mathbb{Z}_{c}\times\mathbb{Z}_{ck}$ for some integers $c\geq 2$ and
$k\geq 1$ that has an $s$-spanning set for some positive integer $s$. If $c$
is an odd prime, then $G$ is $s$-regular.

Proof. First, we prove that $G$ has an $s$-spanning set of the form
$A=\\{(a,u),(b,v)\\}$ for some $a,b\in\mathbb{Z}_{c}$ and
$u,v\in\mathbb{Z}_{ck}$ with $u\not\equiv v$ mod $c$. Note that if
$\\{(a,u),(b,v)\\}$ is an $s$-spanning set, then so is $\\{(a,u),(-b,-v)\\}$,
so if our claim were false, then we would have $u\equiv v\equiv-v$ mod $c$ for
each $s$-spanning set $A=\\{(a,u),(b,v)\\}$ of $G$. But since $c$ is an odd
prime, that would imply that $u\equiv v\equiv 0$ mod $c$, and that is a
contradiction with $A$ being a spanning set.

Suppose then that $A=\\{(a,u),(b,v)\\}$ is an $s$-spanning in $G$ with
$u\not\equiv v$ mod $c$. We will prove that, in this case,
$B=\\{(1,u),(1,v)\\}$ is also an $s$-spanning in $G$.

Let

$I(s)=\\{(\lambda_{1},\lambda_{2})\mid\lambda_{1},\lambda_{2}\in\mathbb{Z},|\lambda_{1}|+|\lambda_{2}|\leq
s\\},$

and consider the function $f:I(s)\rightarrow G$, given by

$(\lambda_{1},\lambda_{2})\mapsto\lambda_{1}(a,u)+\lambda_{2}(b,v).$

Since $A$ is an $s$-spanning set, $f$ is surjective; we can thus find a subset
$I_{0}$ of $I(s)$ on which the restriction $f_{0}:I_{0}\rightarrow G$ of $f$
is bijective. We will prove that $g_{0}:I_{0}\rightarrow G$, given by

$(\lambda_{1},\lambda_{2})\mapsto\lambda_{1}(1,u)+\lambda_{2}(1,v)$

is also bijective. It suffices to show that $g_{0}$ is injective. In order to
do so, let us assume that the equation

$\lambda_{1}(1,u)+\lambda_{2}(1,v)=\mu_{1}(1,u)+\mu_{2}(1,v)$

holds in $G$ for some $(\lambda_{1},\lambda_{2})$ and $(\mu_{1},\mu_{2})$ of
$I_{0}$. We then have

$\displaystyle\lambda_{1}+\lambda_{2}\equiv\mu_{1}+\mu_{2}\;\text{mod}\;c,$
(3)

and

$\displaystyle\lambda_{1}u+\lambda_{2}v\equiv\mu_{1}u+\mu_{2}v\;\text{mod}\;ck.$
(4)

From (4) we find that

$(\lambda_{1}+\lambda_{2})u+\lambda_{2}(v-u)\equiv(\mu_{1}+\mu_{2})u+\mu_{2}(v-u)\;\text{mod}\;ck;$

considering (3), this gives us that

$\lambda_{2}(v-u)\equiv\mu_{2}(v-u)\;\text{mod}\;c.$

Therefore, since $c$ is prime and $u\not\equiv v$ mod $c$, we get that
$\lambda_{2}\equiv\mu_{2}$ mod $c.$ From (3) then we get
$\lambda_{1}\equiv\mu_{1}$ mod $c$, and thus

$\displaystyle\lambda_{1}a+\lambda_{2}b\equiv\mu_{1}a+\mu_{2}b\;\text{mod}\;c.$
(5)

But $f_{0}$ is an injection, so (5) and (4) can hold together only when
$(\lambda_{1},\lambda_{2})=(\mu_{1},\mu_{2})$. This completes our proof.
$\Box$

###### Theorem 3.2

Let $G=\mathbb{Z}_{c}\times\mathbb{Z}_{ck}$ for some integers $c\geq 2$ and
$k\geq 1$, and suppose that $G$ is $s$-regular for some positive integer $s$.

  1. 1.

If $c$ is even, then $|G|\leq 2s^{2},$ and equality may occur only when
$s\equiv 0$ mod $c$.

  2. 2.

If $c$ is odd and

     * •

$s\equiv 0$ mod $c$, then $|G|\leq 2s^{2}+s$;

     * •

$s\equiv(c-1)/2$ mod $c$, then $|G|\leq 2s^{2}+2s-(c^{2}-1)/2;$

     * •

$c=3$ and $s\equiv 2$ mod $3$, then $|G|\leq 2s^{2}+1$.

In all other cases, $|G|<2s^{2}$.

Proof. Let $A=\\{(1,u),(1,v)\\}$ be an $s$-spanning set in $G$, in which case

$\\{(\lambda_{1}+\lambda_{2},\lambda_{1}u+\lambda_{2}v)\mid(\lambda_{1},\lambda_{2})\in\mathbb{Z}^{2},|\lambda_{1}|+|\lambda_{2}|\leq
s\\}=G.$

Recall from Lemma 2.1 that, for a given integer $h$ with $|h|\leq s$, the
number $\Lambda(h)$ of ordered pairs
$(\lambda_{1},\lambda_{2})\in\mathbb{Z}^{2}$ with
$|\lambda_{1}|+|\lambda_{2}|\leq s$ and $\lambda_{1}+\lambda_{2}=h$ equals
$s+1$ when $s$ and $h$ have the same parity, and $s$ otherwise. For each
$i=0,1,\ldots,c-1$, we let $N(i)$ denote the sum of all $\Lambda(h)$ with
$|h|\leq s$ and $h\equiv i$ mod $c$. Since $A$ is an $s$-spanning set in $G$,
we must have

$\\{(\lambda_{1}+\lambda_{2},\lambda_{1}u+\lambda_{2}v)\mid(\lambda_{1},\lambda_{2})\in\mathbb{Z}^{2},|\lambda_{1}|+|\lambda_{2}|\leq
s\\}=\mathbb{Z}_{c}\times\mathbb{Z}_{ck},$

and thus for each $i=0,1,\ldots,c-1$,

$\\{(\lambda_{1}+\lambda_{2},\lambda_{1}u+\lambda_{2}v)\mid(\lambda_{1},\lambda_{2})\in\mathbb{Z}^{2},|\lambda_{1}|+|\lambda_{2}|\leq
s,\lambda_{1}+\lambda_{2}\equiv
i\;\mbox{mod}\;c\\}=\\{i\\}\times\mathbb{Z}_{ck}.$

In particular,

$\displaystyle|G|\leq cN(i)$ (6)

must hold for each $i=0,1,\ldots,c-1$. We now inspect (6) in several cases.

When $c$ is even, we consider $i=r+1$, where $r$ is the nonnegative remainder
of $s$ mod $c$ (with $r+1$ understood to be $0$ when $r=c-1$). To compute
$N(r+1)$, we need to identify the values of $h$ for which $|h|\leq s$ and
$h\equiv r+1$ mod $c$, and these values are $h=s+1-jc$ for
$j=1,2,\ldots,j_{\mathrm{max}}$, where

$j_{\mathrm{max}}=\lfloor(2s+1)/c\rfloor=(2s-2r)/c+\lfloor(2r+1)/c\rfloor.$

Note that $\lfloor(2r+1)/c\rfloor$ equals $0$ when $0\leq r\leq c/2-1$, and
$1$ when $c/2\leq r\leq c-1$. Since $s$ and $h=s+1-jc$ have different parity,
and thus $\Lambda(h)=s$ for each $j$, we have

$cN(r+1)=\left(2s-2r+c\lfloor(2r+1)/c\rfloor\right)s,$

and we see that this quantity is at most $2s^{2}$, with equality when $r=0$ or
$r=c/2$. But $|G|$ must be divisible by $c^{2}$, so $|G|=2s^{2}$ may only
occur when $r=0$.

When $c$ is odd and $r\leq c-2$, we consider $i=r+2$ (where $r+2$ is
understood to be $0$ when $r=c-2$). In this case the values of $h$ with
$|h|\leq s$ and $h\equiv r+2$ mod $c$ are $h=s+2-tc$ for
$t=1,2,\ldots,t_{\mathrm{max}}$, where

$t_{\mathrm{max}}=\lfloor(2s+2)/c\rfloor=(2s-2r)/c+\lfloor(2r+2)/c\rfloor.$

Note that $\lfloor(2r+2)/c\rfloor$ equals $0$ when $0\leq r\leq(c-3)/2$, and
$1$ when $(c-1)/2\leq r\leq c-2$. Since $s$ and $h=s+2-tc$ have the same
parity when $t$ is even and different parity when $t$ is odd, we have

$cN(r+2)=\left(s-r+c\lfloor(2r+2)/c\rfloor\right)s+(s-r)(s+1).$

We consider four subcases. For $r=0$, this quantity equals $2s^{2}+s$, and
when $1\leq r\leq(c-3)/2$, it equals $(s-r)s+(s-r)(s+1)$, which is less than
$2s^{2}$. When $(c+1)/2\leq r\leq c-2$, we have

$cN(r+2)=\left(s-r+c\right)s+(s-r)(s+1)=(s-r)(2s+1)+cs\leq\left(s-(c+1)/2\right)(2s+1)+cs,$

which again is less than $2s^{2}$. Our last subcase is when $r=(c-1)/2$, when
we have

$cN(r+2)=\left(s-(c-1)/2\right)(2s+1)+cs=2s^{2}+2s-(c-1)/2.$

Observe that when $s\equiv(c-1)/2$ mod $c$, then $2s^{2}+2s-(c^{2}-1)/2\equiv
0$ mod $c^{2}$, and since $|G|$ must be divisible by $c^{2}$, this is the
largest it can then be.

Our final case is when $c$ is odd and $r=c-1$; we consider $i=0$. In this case
the values of $h$ with $|h|\leq s$ and $h\equiv 0$ mod $c$ are $h=s+1-zc$ for
$z=1,2,\ldots,z_{\mathrm{max}}$, where

$z_{\mathrm{max}}=\lfloor(2s+1)/c\rfloor=(2s+2-c)/c.$

Note that $z_{\mathrm{max}}$ is odd. Since $s$ and $h=s+1-zc$ have the same
parity when $z$ is odd and different parity when $z$ is even, we have

$N(0)=(z_{\mathrm{max}}-1)/2\cdot s+(z_{\mathrm{max}}+1)/2\cdot(s+1),$

and thus

$cN(0)=\left(s+1-c\right)s+\left(s+1\right)(s+1)=2s^{2}-(c-3)s+1,$

which is less than $2s^{2}$, unless $c=3$. This completes our proof. $\Box$

###### Theorem 3.3

For $s\geq 3$, every $s$-regular abelian group of rank 2 has order at most
$2s^{2}+2s-4$, with equality if, and only if, it is isomorphic to
$\mathbb{Z}_{3}\times\mathbb{Z}_{(2s^{2}+2s-4)/3}$ for some $s\equiv 1$ mod
$3$.

Proof. Suppose that $G=\mathbb{Z}_{c}\times\mathbb{Z}_{ck}$ is an $s$-regular
group of order at least $2s^{2}+2s-4$ for some integers $s\geq 3$, $c\geq 2$,
and $k\geq 1$. Then, by Theorem 3.2, $c$ is odd; furthermore,

  * •

$s\equiv 0$ mod $c$, and $2s^{2}+2s-4\leq c^{2}k\leq 2s^{2}+s$, or

  * •

$s\equiv(c-1)/2$ mod $c$, and $2s^{2}+2s-4\leq c^{2}k\leq
2s^{2}+2s-(c^{2}-1)/2.$

Since the first possibility leads to no solutions, our claim is established.
$\Box$

##  4 The groups $\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$

In this section we provide some results for $s$-spanning sets in groups of the
form $\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$, including the exact value of the
largest possible order of such groups. Our general conjecture is as follows.

###### Conjecture 4.1

Let $k$ and $s$ be positive integers. The group
$\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$ has an $s$-spanning set of size $2$ if,
and only if, one of the following holds:

  * •

$s$ is odd and $k\leq(s^{2}-1)/2$;

  * •

$s$ is even and $k\leq(s^{2}-s)/2$;

  * •

$s\equiv 0$ mod $4$, $k$ is even, and $(s^{2}-s)/2<k\leq s^{2}/2$; or

  * •

$s\equiv 2$ mod $4$, $k\equiv 2$ mod $4$, and $(s^{2}-s)/2<k\leq s^{2}/2$.

###### Theorem 4.2

Let $k$ and $s$ be positive integers. If the group
$G=\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$ has an $s$-spanning set of size 2,
then it has order at most $2s^{2}$.

Proof. If $G$ is $s$-regular, that is, it has an $s$-spanning set of the form
$\\{(1,u),(1,v)\\}$ for some $u,v\in\mathbb{Z}_{2k}$, then our claim follows
from Theorem 3.2. Since no set of the form $\\{(0,u),(0,v)\\}$ can be a
spanning set, we can suppose that $G$ has an $s$-spanning set of the form
$A=\\{(0,u),(1,v)\\}$ for some $u,v\in\mathbb{Z}_{2k}$. Therefore, each
element in $\langle A\rangle_{s}$ has the form
$(\lambda_{2},\lambda_{1}u+\lambda_{2}v)$ for some integers $\lambda_{1}$ and
$\lambda_{2}$ with $|\lambda_{1}|+|\lambda_{2}|\leq s$.

Clearly, if $u$ and $v$ are both even, then $A$ can only generate elements of
$G$ with an even second component, and thus it cannot be a spanning set.

Now if $u$ and $v$ are both odd, then $\lambda_{1}u+\lambda_{2}v$ has the same
parity as $\lambda_{1}+\lambda_{2}$ does. According to Lemma 2.1, the parity
of $\lambda_{1}+\lambda_{2}$ differs from the parity of $s$ exactly $s^{2}$
times (and agrees with it $(s+1)^{2}$ times) within
$|\lambda_{1}|+|\lambda_{2}|\leq s$. This means that $A$ cannot generate more
than $s^{2}$ elements in $\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$ whose second
coordinate has different parity than $s$ does, so if it is an $s$-spanning
set, then we must have $|G|\leq 2s^{2}$.

Next, if $u$ is even and $v$ is odd, then for each element of $G$ that $A$
generates, its two components will have the same parity (namely, the parity of
$\lambda_{2}$). Therefore, $A$ cannot be a spanning set.

Finally, we prove that if $A=\\{(0,u),(1,v)\\}$ is an $s$-spanning set of $G$
with $u$ odd and $v$ even, then $A^{\prime}=\\{(1,u),(1,v)\\}$ is an
$s$-spanning set as well, meaning that $G$ is $s$-regular and thus has order
at most $2s^{2}$ by Theorem 3.2.

Let $a\in\mathbb{Z}_{2}$ and $b\in\mathbb{Z}_{2k}$. Since $\langle
A\rangle_{s}=G$, there are integers $\lambda_{1}$ and $\lambda_{2}$ with
$|\lambda_{1}|+|\lambda_{2}|\leq s$ for which

$\lambda_{1}(0,u)+\lambda_{2}(1,v)=(\lambda_{2},\lambda_{1}u+\lambda_{2}v)=(a^{\prime},b),$

where $a^{\prime}=a$ if $b$ is even and $a^{\prime}=1+a$ when $b$ is odd.
Observe that when $b$ is even, then (since $u$ is odd and $v$ is even),
$\lambda_{1}$ must be even, and thus

$\lambda_{1}(1,u)+\lambda_{2}(1,v)=(\lambda_{1}+\lambda_{2},\lambda_{1}u+\lambda_{2}v)=(\lambda_{2},\lambda_{1}u+\lambda_{2}v)=(a^{\prime},b)=(a,b).$

Similarly, when $b$ is odd, $\lambda_{1}$ must be odd, and

$\lambda_{1}(1,u)+\lambda_{2}(1,v)=(1+a^{\prime},b)=(a,b).$

Therefore, $\langle A^{\prime}\rangle_{s}=G$, as claimed, which completes our
proof. $\Box$

###### Theorem 4.3

Let $k$ and $s$ be positive integers with $s$ odd. Then the group
$G=\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$ has an $s$-spanning set of size 2 if,
and only if, $k\leq(s^{2}-1)/2$.

Proof. The fact that $k\leq(s^{2}-1)/2$ is necessary follows from Theorem 4.2.
We may also recall that Theorem 2.4, applied with $c=2$ and odd $s$, yields
that with $u=(s-1)/2$ and $v=(s+1)/2$, the set $\\{(1,u),(1,v)\\}$ is an
$s$-spanning set in $\mathbb{Z}_{2}\times\mathbb{Z}_{s^{2}-1}$. This takes
care of the case $k=(s^{2}-1)/2$.

Our goal below is to show that the same values of $u=(s-1)/2$ and $v=(s+1)/2$
as integers provide an $s$-span for the set
$\mathbb{Z}_{2}\times\\{0,1,2,\ldots,k\\}$ when $k\leq(s^{2}-3)/2$. (Here we
consider the operation in the first component in $\mathbb{Z}_{2}$ and in the
second component in $\mathbb{Z}$.) By symmetry, we can then conclude that
$\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$ has an $s$-spanning set of size 2 for
$k\leq(s^{2}-3)/2$ as well.

We thus need to establish that for every $a\in\mathbb{Z}_{2}$ and
$b\in\\{0,1,2,\ldots,k\\}$, there are integers $\mu_{1}$ and $\mu_{2}$ with
$|\mu_{1}|+|\mu_{2}|\leq s$ for which

$\displaystyle\mu_{1}(1,(s-1)/2)+\mu_{2}(1,(s+1)/2)=(a,b)$ (7)

holds in $\mathbb{Z}_{2}\times\\{0,1,2,\ldots,k\\}$.

Let us recall from the proof of Theorem 2.4 (applied again for odd $s$ and
$c=2$) that there are integers $\lambda_{1}\in[-s,s+1]$ and
$\lambda_{2}\in[0,s-2]$ for which one of the following two sets of conditions
holds:

  1. (i)

$|\lambda_{1}|+|\lambda_{2}|\leq s$, and equation (7) is satisfied in
$\mathbb{Z}_{2}\times\mathbb{Z}_{s^{2}-1}$ with
$(\mu_{1},\mu_{2})=(\lambda_{1},\lambda_{2})$; or

  2. (ii)

$|\lambda_{1}|+|\lambda_{2}|\geq s+1$, and equation (7) is satisfied in
$\mathbb{Z}_{2}\times\mathbb{Z}_{s^{2}-1}$ with
$(\mu_{1},\mu_{2})=(\lambda_{1}-\epsilon(s+1),\lambda_{2}-s+1)$, where
$\epsilon$ equals $1$ when $\lambda_{1}\geq 0$, and $-1$ when $\lambda_{1}<0$.

To complete our proof, we will show that, considering

$B=\mu_{1}(s-1)/2+\mu_{2}(s+1)/2$

as an integer (rather than as an element of $\mathbb{Z}_{s^{2}-1}$), we have
$B=b$. Since $0\leq b\leq k<(s^{2}-1)/2$, it suffices to prove that

$\displaystyle-(s^{2}-1)/2\leq B<s^{2}-1.$ (8)

In case (i), we have $\mu_{1}=\lambda_{1}\in[-s,s+1]$ and
$\mu_{2}=\lambda_{2}\in[0,s-2]$, from which (8) readily follows. For case
(ii), we find that

$|\mu_{1}|+|\mu_{2}|=(s+1-|\lambda_{1}|)+(s-1-|\lambda_{2}|)\leq 2s-(s+1)=s-1$

and thus

$|B|=|\mu_{1}(s-1)/2+\mu_{2}(s+1)/2|\leq|\mu_{1}|(s-1)/2+|\mu_{2}|(s+1)/2\leq(|\mu_{1}|+|\mu_{2}|)(s+1)/2\leq(s^{2}-1)/2,$

so (8) holds again. This completes our proof. $\Box$

###### Theorem 4.4

Let $k$ and $s$ be positive integers with $s$ even. Then the group
$G=\mathbb{Z}_{2}\times\mathbb{Z}_{2k}$ has an $s$-spanning set of size 2 if
$k\leq(s^{2}-s)/2$.

Proof. Like in the proof of Theorem 4.3, we will prove that there are integers
$u$ and $v$ for which the $s$-span of $A=\\{(1,u),(1,v)\\}$ contains the set
$\mathbb{Z}_{2}\times\\{0,1,2,\ldots,k\\}$. (Here too we consider the
operation in the first component in $\mathbb{Z}_{2}$ and in the second
component in $\mathbb{Z}$.)

Since $s-1$ is odd, from the proof of Theorem 4.3 we see that with $u=(s-2)/2$
and $v=s/2$, $\langle A\rangle_{s-1}$ contains the set
$\mathbb{Z}_{2}\times\\{0,1,2,\ldots,k\\}$ when $k\leq((s-1)^{2}-1)/2$.
Therefore, $(1,v)+\langle A\rangle_{s-1}$ contains
$\mathbb{Z}_{2}\times\\{s/2,s/2+1,\ldots,s/2+k\\}$ for each
$k\leq(s^{2}-2s)/2$, and since $(1,v)+\langle A\rangle_{s-1}\subseteq\langle
A\rangle_{s}$, our claim follows for all $k\leq(s^{2}-s)/2$. $\Box$

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