the following was an old conjecture of graham  . conjecture  1.1let 
 cp 

be the cyclic group of order 
 p 
 prime and let 
 s 
 be a sequence over 
 cp 
 of length 
 p 
 .
if all ( nontrivial ) zero - sum subsequences of 
 s 
 are of the same length , then the number of distinct terms in 
 s 
 is at most 2 .
let 
 cp 
 be the cyclic group of order 
 p 
 prime and let 
 s 
 be a sequence over 
 cp 
 of length 
 p 

. if all ( nontrivial ) zero - sum subsequences of 
 s 
 are of the same length , then the number of distinct terms in 
 s 
 is at most 2 . in 1976 ,
erds and szemerdi gave a proof of the conjecture for sufficiently large primes p  .
however , the proof was complicated enough that the details for small primes were never worked out . both in the paper of erds and szemerdi and in a later survey by erds and graham  , the complexity of the proof was lamented .
recently , a new proof , valid even for non - primes , was given by gao et  al . 
, using savchev and chen s recently proved structure theorem for zero - sum free sequences of long length in the cyclic group cn  .
however , as savchev and chen s result is fairly involved , they did not believe it to be the simple proof sought by erds , graham and szemerdi . in this paper
, we give a short proof to the original conjecture of graham that uses only the cauchy  davenport theorem and pigeonhole principle  , . since the proof of the cauchy  davenport theorem ( known since 1813  ) is elementary and requires only a paragraph , our proof may perhaps qualify as simple . replacing the use of the cauchy  davenport theorem with the devos  goddyn  mohar theorem   ( alternatively , the partition theorem from  ,   could be used instead of devos  goddyn  mohar ) , we obtain an alternate proof , albeit not as simple , of the non - prime case . with only a little added effort
, our method naturally yields an exhaustive list detailing the precise structure of s and shows that the result holds in an arbitrary finite abelian group , though the only additional group for which the hypotheses are non - void is c2c2 m .
we state the main theorem in section  3 , after introducing modern notation for sumsets , sequences and subsequence sums .
we follow the notation of  , ,   and   concerning sumsets , sequences and subsequence sums . for the convenience of the reader less familiar with this notation
let g be an abelian group , and let a , bg be nonempty subsets .
we let g+a={g+aaa } and let ra , b(g ) denote the number of representations of g = a+b as a sum with aa and bb .
the stabilizer of a is h(a):={ggg+a = a}. the order of an element gg is denoted by ord(g ) , and we use h : gg / h to denote the natural homomorphism modulo h. we use cn to denote the cyclic group of order n. we let f(g ) denote the free abelian monoid with basis g written multiplicatively .
the elements of f(g ) are then just multi - sets over g , but following long standing tradition , we refer to the sf(g ) as sequences .
we write sequences sf(g ) in the form s = s1sr=gggvg(s),wherevg(s)0  and  sig .
we call |s|r=ggvg(s ) the length of s , and vg(s)n0 the  multiplicity  of g in s. the support of s is supp(s){ggvg(s)>0}. a sequence s1 is called a subsequence of s if s1|s in f(g ) ( equivalently , vg(s1)vg(s ) for all gg ) , and in such case , ss11 or s11s denotes the subsequence of s obtained by removing all terms from s1
. we let h(s)max{vg(s)gg } denote the maximum multiplicity of a term of s. given any map :gg , we extend  to a map of sequences , :f(g)f(g ) , by letting (s)(s1)(sr ) .
if s = s1srf(g ) , with sig , then the sum of s is (s)i=1rsi=ggvg(s)g .
we adapt the convention that the sum of the trivial / empty sequence is zero .
we follow the usual notation for the set of subsequence sums : n(s)={(t)t|s  and  |t|=n}n(s)=i=1ni(s)andn(s)=i = n|s|i(s)and(s)=|s|(s ) . for a finite abelian group g
, we define the davenport constant d(g ) to be the minimal integer such that any sf(g ) with |s|d(g ) has 0(s ) .
a basic argument shows that d(g)|g| ( see [ 9 ,  propositions 5.1.4 ] ) .
we need the following result ( see  [ 9 ,  theorem 5.2.10 ; lemma 5.2.9 ] and also  [ 14 ,  lemma 2.1 ] ) .
proposition  2.1(ii ) is a simple consequence of the pigeonhole principle , and we will only use proposition  2.1(i ) in the trivial case |b|=k=2 .
proposition  2.1let 
 g 
 be an abelian group with 
 a , bg 
 finite and nonempty:(i)if 
 |a+b||a|+|b|k , then 
 ra , b(x)k 
 for all 
 xa+b 
 ; ( ii)if 
 g 
 is finite and 
 |a|+|b||g|+k , then 
 ra , b(x)k 
 for all 
 xg .
let 
 g 
 be an abelian group with 
 a , bg 
 finite and nonempty:(i)if 
 |a+b||a|+|b|k , then 
 ra , b(x)k 
 for all 
 xa+b 
 ; ( ii)if 
 g 
 is finite and 
 |a|+|b||g|+k , then 
 ra , b(x)k 
 for all 
 xg . if 
 |a+b||a|+|b|k , then 
 ra , b(x)k 
 for all 
 xa+b 
 ; if 
 g 
 is finite and 
 |a|+|b||g|+k , then 
 ra , b(x)k 
 for all 
 xg .
theorem  2.2let 
 g 
 be an abelian group , let 
 sf(g ) 
 be a sequence , and let 
 nz+ 
 with 
 n|s| 
 .
let 
 g 
 be an abelian group , let 
 sf(g ) 
 be a sequence , and let 
 nz+ 
 with 
 n|s| 
 .
if 
 h = h(n(s ) ) , then(1)|n(s)|(gg / hmin{n , vg(h(s))}n+1)|h| . a particular case of the ( general ) devos  goddyn  mohar theorem is the much simpler cauchy  davenport theorem  , , , .
let p be prime and let aicp , for i=1,  ,n , be nonempty . then |i=1nai|min{i=1n|ai|n+1,p}.
let g be an abelian group , and let a , bg be nonempty subsets .
we let g+a={g+aaa } and let ra , b(g ) denote the number of representations of g = a+b as a sum with aa and bb .
the stabilizer of a is h(a):={ggg+a = a}. the order of an element gg is denoted by ord(g ) , and we use h : gg / h to denote the natural homomorphism modulo h. we use cn to denote the cyclic group of order n.
we let f(g ) denote the free abelian monoid with basis g written multiplicatively . the elements of f(g ) are then just multi - sets over g , but following long standing tradition
we write sequences sf(g ) in the form s = s1sr=gggvg(s),wherevg(s)0  and  sig .
we call |s|r=ggvg(s ) the length of s , and vg(s)n0 the  multiplicity  of g in s. the support of s is supp(s){ggvg(s)>0}. a sequence s1 is called a subsequence of s if s1|s in f(g ) ( equivalently , vg(s1)vg(s ) for all gg ) , and in such case , ss11 or s11s denotes the subsequence of s obtained by removing all terms from s1 .
we let h(s)max{vg(s)gg } denote the maximum multiplicity of a term of s. given any map :gg , we extend  to a map of sequences , :f(g)f(g ) , by letting (s)(s1)(sr ) .
if s = s1srf(g ) , with sig , then the sum of s is (s)i=1rsi=ggvg(s)g .
we adapt the convention that the sum of the trivial / empty sequence is zero .
we follow the usual notation for the set of subsequence sums : n(s)={(t)t|s  and  |t|=n}n(s)=i=1ni(s)andn(s)=i = n|s|i(s)and(s)=|s|(s ) .
for a finite abelian group g , we define the davenport constant d(g ) to be the minimal integer such that any sf(g ) with |s|d(g ) has 0(s ) .
a basic argument shows that d(g)|g| ( see [ 9 ,  propositions 5.1.4 ] ) .
we need the following result ( see  [ 9 ,  theorem 5.2.10 ; lemma 5.2.9 ] and also  [ 14 ,  lemma 2.1 ] ) .
proposition  2.1(ii ) is a simple consequence of the pigeonhole principle , and we will only use proposition  2.1(i ) in the trivial case |b|=k=2 .
proposition  2.1let 
 g 
 be an abelian group with 
 a , bg 
 finite and nonempty:(i)if 
 |a+b||a|+|b|k , then 
 ra , b(x)k 
 for all 
 xa+b 
 ; ( ii)if 
 g 
 is finite and 
 |a|+|b||g|+k , then 
 ra , b(x)k 
 for all 
 xg .
let 
 g 
 be an abelian group with 
 a , bg 
 finite and nonempty:(i)if 
 |a+b||a|+|b|k , then 
 ra , b(x)k 
 for all 
 xa+b 
 ; ( ii)if 
 g 
 is finite and 
 |a|+|b||g|+k , then 
 ra , b(x)k 
 for all 
 xg . if 
 |a+b||a|+|b|k , then 
 ra , b(x)k 
 for all 
 xa+b 
 ; if 
 g 
 is finite and 
 |a|+|b||g|+k , then 
 ra , b(x)k 
 for all 
 xg .
theorem  2.2let 
 g 
 be an abelian group , let 
 sf(g ) 
 be a sequence , and let 
 nz+ 
 with 
 n|s| 
 .
let 
 g 
 be an abelian group , let 
 sf(g ) 
 be a sequence , and let 
 nz+ 
 with 
 n|s| 
 .
if 
 h = h(n(s ) ) , then(1)|n(s)|(gg / hmin{n , vg(h(s))}n+1)|h| . a particular case of the ( general ) devos  goddyn  mohar theorem is the much simpler cauchy  davenport theorem  , , , .
let p be prime and let aicp , for i=1,  ,n , be nonempty . then |i=1nai|min{i=1n|ai|n+1,p}.
lemma  3.1let 
 g 
 be an abelian group , let 
 gg , and let 
 rf(g ) 
 be nontrivial with(2)(r){g,2g, 
let 
 g 
 be an abelian group , let 
 gg , and let 
 rf(g ) 
 be nontrivial with(2)(r){g,2g, 
proofthe result is clear when |r|2 , so we may assume that |r|3 . in view of ( 2 ) and |r|ord(g)1 , we have 0(r ) .
note , since |r|ord(g)1 , that ( 3 ) shows h0 as well . from 0(r ) and ( 2 ) ( note if r|rh1 with (r)=(r ) , then (rr1)=0 ; hence (r)(rh1 ) ) , we have (rh1)({g,2g,  ,|r|g}{(r)})({g,2g, 
h{g,0 } , 0{g,2g,  ,|r|g } and (r)=|r|g imply that |(rh1)|<|rh1|=|r|1 . as a result
, |(rh1){0}|=|i=1|r|1{0,gi}||r|1 , where rh1=g1g|r|1 with gisupp(r)(r)g{0}. hence proposition  2.1(i ) ( applied to the partial sums i=1j1{0,gi}+{0,gj } ) implies that every element of i=1|r|1{0,gi } has at least two representations , contradicting that 0(r ) . 
note , since |r|ord(g)1 , that ( 3 ) shows h0 as well . from 0(r ) and ( 2 ) ( note if r|rh1 with (r)=(r ) , then (rr1)=0 ; hence (r)(rh1 ) ) , we have (rh1)({g,2g,  ,|r|g}{(r)})({g,2g, 
h{g,0 } , 0{g,2g,  ,|r|g } and (r)=|r|g imply that |(rh1)|<|rh1|=|r|1 . as a result , |(rh1){0}|=|i=1|r|1{0,gi}||r|1 , where rh1=g1g|r|1 with gisupp(r)(r)g{0}. hence proposition  2.1(i ) ( applied to the partial sums i=1j1{0,gi}+{0,gj } ) implies that every element of i=1|r|1{0,gi } has at least two representations , contradicting that 0(r ) . 
 the next two lemmas will help with the detailed characterization of s. lemma  3.2let 
 g , hcn 
 and let 
 sf(cn ) 
 with 
 s = glhnl 
 and 
 lnl1 
 .
suppose 
 g 
 is a generator and(4)(hnl)={g,2g,  ,(nl1)g}{b0},for some 
 b0cn 
 . if there is a unique 
 r[1,n ] 
 such that 
 0r(s ) , then either 
 s = gn1h 
 or else 
 n 
 is odd , 
 h = n+12 g 
 and 
 s = gn2h2 .
let 
 g , hcn 
 and let 
 sf(cn ) 
 with 
 s = glhnl 
 and 
 lnl1 
 .
suppose 
 g 
 is a generator and(4)(hnl)={g,2g,  ,(nl1)g}{b0},for some 
 b0cn 
 . if there is a unique 
 r[1,n ] 
 such that 
 0r(s ) , then either 
 s = gn1h 
 or else 
 n 
 is odd , 

h = n+12 g 
 and 
 s = gn2h2 . proofthe cases nl2 and l1 are easily verified , so we may assume that 3nln2 and thus hg ( else either there are two disjoint zero - sums of length 2 or s = gn1h = gn )
. now ( 4 ) implies (hnl)={h,2h,  ,(nl)h}={g,2g,  ,(nl1)g}{b0 } , for some b0cn .
thus { h,2h,  ,(nl)h } contains an arithmetic progression of difference gh and length nl12 .
if nl4 , then it is easily seen , in view of the hypothesis nln2 , that { h,2h,  ,(nl)h } can not contain an arithmetic progression of length nl1 and difference gh . on the other hand , if nl=3 , then this could only be possible if g=2h , and this final case can be eliminated by individual consideration , completing the proof . 
 the cases nl2 and l1 are easily verified , so we may assume that 3nln2 and thus hg ( else either there are two disjoint zero - sums of length 2 or s = gn1h = gn )
. now ( 4 ) implies (hnl)={h,2h,  ,(nl)h}={g,2g,  ,(nl1)g}{b0 } , for some b0cn .
thus { h,2h,  ,(nl)h } contains an arithmetic progression of difference gh and length nl12 .
hence , if nl4 , then it is easily seen , in view of the hypothesis nln2 , that { h,2h,  ,(nl)h } can not contain an arithmetic progression of length nl1 and difference gh . on the other hand , if nl=3 , then this could only be possible if g=2h , and this final case can be eliminated by individual consideration , completing the proof .   lemma  3.3let 
 g 
 be an abelian group of order 
 n 
 even , let 
 g , hg 
 with 
 ord(g)=n2 
 and 
 hg , and let 
 sf(g ) 
 with 
 s = glhnl , 
 nl2 
 and 
 ln2 
 . if 
 n2[1,n ] 
 is the unique integer 
 r 
 such that 
 0r(s ) , then 
 nl 
 is odd , 
 hg 
 and 
 2h=2 g .
let 
 g 
 be an abelian group of order 
 n 
 even , let 
 g , hg 
 with 
 ord(g)=n2 
 and 
 hg , and let 
 sf(g ) 
 with 
 s = glhnl , 
 nl2 
 and 
 ln2 
 .
if 
 n2[1,n ] 
 is the unique integer 
 r 
 such that 
 0r(s ) , then 
 nl 
 is odd , 
 hg 
 and 
 2h=2 g .
proofsince hg , ln2 and ord(g)=n2[1,n ] is the unique integer r such that 0r(s ) , we conclude that hg. however , noting that 2hg ( since g has index 2 ) , we likewise see that we must have 2h=2 g ( in view of nl2 ) , else the uniqueness of n2=ord(g ) is again contradicted .
consequently , the sum of any n2-terms of s using an even number of terms from hnl has sum zero . as a result , if nl is even , then there are two disjoint zero - sum subsequences of length n2 , contradicting the uniqueness of n2 , and completing the proof . 
 since hg , ln2 and ord(g)=n2[1,n ] is the unique integer r such that 0r(s ) , we conclude that hg. however , noting that 2hg ( since g has index 2 ) , we likewise see that we must have 2h=2 g ( in view of nl2 ) , else the uniqueness of n2=ord(g ) is again contradicted .
consequently , the sum of any n2-terms of s using an even number of terms from hnl has sum zero . as a result ,
if nl is even , then there are two disjoint zero - sum subsequences of length n2 , contradicting the uniqueness of n2 , and completing the proof . 
 next , we state and prove the main result . in the remark that follows the proof of theorem  3.4 ,
we explain how the proof can be simplified in the case g = cp with p prime , including the use of the cauchy  davenport theorem in place of devos  goddyn  mohar .
also , though we state the theorem for an arbitrary finite abelian group , most non - cyclic cases have no sequences satisfying the hypotheses ( since 2d(g)|g| holds for most non - cyclic groups  [ 9 ,  theorem 5.5.5 ] . )
theorem  3.4let 
 g 
 be a finite abelian group of order 
 n 
 and let 
 sf(g ) 
 with 
 |s|=n 
 .
suppose there is a unique 
 r[1,n ] 
 such that 
 0r(s ) 
 .
then 
 |supp(s)|2.if 
 g 
 is non - cyclic , then 
 g=hgc2c2 m , 
 r = n2=2 m 
 ands = gn1g  or  s = gn/2+x(h+g)n/2x  or  s = gn/2+x(h+n+44g)n/2x , where 
 gg , 
 h , ggg , 
 ord(g)=n2 , 
 ord(h)=2 
 and 
 x[1,n21 ] 
 is odd.if 
 g 
 is cyclic , then there exists a generator 
 ggcn 
 such that eithers = gn1g  or  s=(2g)n1g,for some 
 gg 
 or 
 gg2g 
 ; or 
 n 
 is odd , 
 r = n+12 
 ands = gn2(n+12g)2;or 
 n2mod4 , 
 r = n2 
 ands=(2g)n/2+x(n+42g)n/2x , where 
 x[0,n21 ] 
 is even ; or 
 n 
 is even , 
 r = n2 
 ands = gn/2+x(n+22g)n/2x , where 
 x[0,n21 ] 
 with 
 n2x 
 odd . let 
 g 
 be a finite abelian group of order 
 n 
 and let 
 sf(g ) 
 with 
 |s|=n 
 .
. then 
 |supp(s)|2 . if 
 g 
 is non - cyclic , then 
 g=hgc2c2 m , 
 r = n2=2 m 
 ands = gn1g  or  s = gn/2+x(h+g)n/2x  or  s = gn/2+x(h+n+44g)n/2x , where 
 gg , 
 h , ggg , 
 ord(g)=n2 , 
 ord(h)=2 
 and 
 x[1,n21 ] 
 is odd . if 
 g 
 is cyclic , then there exists a generator 
 ggcn 
 such that eithers = gn1g  or  s=(2g)n1g,for some 
 gg 
 or 
 gg2g 
 ; or 
 n 
 is odd , 
 r = n+12 
 ands = gn2(n+12g)2;or 
 n2mod4 , 
 r = n2 
 ands=(2g)n/2+x(n+42g)n/2x , where 
 x[0,n21 ] 
 is even ; or 
 n 
 is even , 
 r = n2 
 ands = gn/2+x(n+22g)n/2x , where 
 x[0,n21 ] 
 with 
 n2x 
 odd . proofrecalling the well - known fact that a zero - sum free subsequence of length |g|1 must be of the form g|g|1 for a generator gg ( this can be proved in a few lines using the trivial case |b|=k=2 in proposition  2.1(i ) ; see also  [ 9 ,  lemma 5.4.2 ] for a slightly more involved proof ) , we see that the cases r=1 and r = n are trivial
observe that(5)0r1(s)=r1(0r2s),(6)0r+1(s)=n(0nr1s)=(s)nr1(0nr1s ) , where we have used for ( 6 ) the fact that m(t)=(t)|t|m(t ) for tf(g ) , which follows in view of the correspondence between r|t and tr1|t.step 1 .
we first show that either ( 7)h(s)max{r , nr+1 } ,  or(8)h(s)max{r , nr}ands = gn/2gn/2withord(g)=ord(g)=n  even  , where gg .
note that if there are distinct g , gsupp(s ) each with multiplicity at least nr , then ( 9 ) implies that n is even with s = gn/2gn/2 and r = n2 . if ord(g)=ord(g)=n , then ( 8) holds , as desired . on the other hand ,
if ( say ) ord(g)n2 , then ord(g)=r = n2 , in which case the proof is easily concluded using lemma  3.3 .
therefore we may assume that there is at most one term with multiplicity at least nr.we apply theorem  2.2 to nr1(0nr1s ) .
let h = h(nr1(0nr1s ) ) . now assuming h(s)nr , it follows , in view of ( 1 ) , ( 6 ) , and since there is at most one term of multiplicity nr , that h is a proper , nontrivial subgroup .
moreover , in view of v0(0nr1s)=nr1n21|g / h|1 , which follows from ( 9 ) , we see that ( 1 ) implies that all but at most |g / h|2 terms of s are from h. letting t|s be the subsequence of all terms not from h , we see that (s)(t)+h .
thus , since |t||g / h|2n21nr1 ( by ( 9 ) ) , it follows , in view of the definition of h , that (s)nr1(0nr1s ) , in contradiction to ( 6 ) .
therefore we may instead assume that ( 9 ) fails , i.e. , ( 10)r1>n21.in this case , we apply theorem  2.2 to r1(0r2s ) .
however , assuming h(s)r1 and repeating the above arguments using ( 5 ) instead of ( 6 ) and using ( 10 ) instead of ( 9 ) , we arrive at the same contradiction .
thus ( 7 ) is established in both cases.factor s = glt , where tf(g ) , and let r|t be a maximal length subsequence ( possibly trivial ) such that (r)=|r|g . in view of ( 7 ) , ( 8) and ( 5 ) , it follows that(11)vg(s)=l = h(s)max{r , nr}n2|t| and 0(t ) ; in particular , 0(r).step 2 .
then it follows in view of ( 11 ) that r = ord(g ) and that g is the only element from hg in supp(s ) ( else we can find a zero - sum of length distinct from r ) . iteratively applying the definition of d(g / h)|g / h| to h(u1sgord(g ) ) , beginning with u trivial , we find a zero - sum mod h subsequence u|sgord(g ) with |u|n|h||g / h|+1 .
adding on an appropriate number of terms from gord(g ) ( note (gord(g))=h ) yields a zero - sum subsequence u|s with |u|n|h||g / h|+2 . if |h|<n2 , then |u|>|h|=r , a contradiction . on the other hand ,
if |h|=n2 , then we obtain the same contradiction unless |u|=n21 , (u)=g and su1gn/2=g0h .
thus , if there is some g0supp(t)h with g0g0 , then swapping g0 for g0 in u yields a new u|sgord(g ) with (u)h and |u|=n21 but (u)g , whence we obtain the contradiction as before .
therefore , we instead see that all terms outside h in supp(s ) are equal to g0 .
however , since all terms inside h in supp(s ) are equal to g , this shows that |supp(s)|2 .
so we henceforth assume that ord(g)=n , in which case gcn is cyclic.since |r||t|rln2=ord(g)2 ( the last inequality holds else the proof is complete , while the other inequalities follow from ( 11 ) ) , and since (r)=|r|g , it follows that ( 12)0{g,2g,  ,rg}(gl),(13)0{(r+1)g,(r+2)g,  ,(l+|r|)g}r+1(glr ) . hence l+|r|ord(g)1=n1 and |r|<|t|=nlr.step 3
next , we show that , when r is nontrivial , there is some(14)hr+1(glr){g,2g,  ,(l+|r|)g}. thus assume for the moment that r is nontrivial .
then , in view of lemma  3.1 and 0(r ) , there is some nontrivial r0|r with (r0){0,g,  ,|r|g}. note (r0)|r|g=(r ) implies |r0|<|r|<|t|r ; thus 1|r0|r2 . if (r0){g,2g,  ,(r|r0|1)g } , then 0r1(glr0 ) , contrary to hypothesis .
,(nr+|r0|)g } , whence l+|r|ord(g)1=n1 and gl|s with lrr|r0|+10 show that either (r0)=(nr+|r0|)g=(|r0|r)g or else ( 14 ) holds , as desired .
however , in the former case , factor r = r0r1 and note that (r1)=(r)(r0)=(|r1|+r)g .
now |r|<|t|<|r1|+r|t|1+rn1 , where the last inequality follows from |t|=nl with lr , whence (r1){0,g,  ,|r|g } ( in view of ord(g)=n ) and so |r1|<|r|<|t|r .
thus 1|r1|r2 , and applying the above arguments with r1 instead of r0 , we establish ( 14 ) unless ( |r1|+r)g=(r1)=(|r1|r)g .
however , the latter case implies 2rg=0 , whence r = n2 with n even.furthermore , by the above work for r0 and r1 , we see that ( 14 ) is established unless ( 15)(r){g,2g,  ,|r|g}{(|r|n2)g } for all nontrivial r|r . applying ( 15 ) to each xsupp(r ) , noting that gsupp(r ) ( in view of r|t ) , and recalling that |r|<|t|r = n2 , we conclude that supp(r){2g,3g,  ,(n21)g}{(n2 + 1)g}. if there are ( n2 + 1)g , xsupp(r ) with x{2g,3g, 
,(n21)g } , then applying ( 15 ) to the sequence x((n2 + 1)g ) yields a contradiction .
therefore we conclude that either ( 16)supp(r)={(n2 + 1)g }  or  supp(r){2g,3g,  ,(n21)g}.noting that n2g = rg=rg and (r1)=(|r1|+r)g , we see that { ( |r1|+r+1)g,(|r1|+r+2)g,  ,(n2)g}(r1)+r|r1|2(gl)r2(glr1 ) .
thus , since |r1|+r+1|r|+l+1 , and in view of ( 13 ) and ord(g)=n , we have g{g,(r1)g}r2(glr)r(glr ) . as a result ( recall |r|<|t|),(17)supp(tr1)={(r1)g}={(n2 + 1)g},(18)(r1)gr2(glr)r(glr )
, else we find a zero - sum of length distinct from r using precisely one term from tr1 ( recall gsupp(t ) in view of the definition of t ) , contrary to hypothesis.by ( 16 ) , ( 17 ) , we discover that supp(r){2g,3g, 
,(n21)g } , else supp(s)={g , n+22 g } with r = n2 , from which the remainder of the proof is easily deduced .
thus , since r = n2 and r is nontrivial , we see that r3 and ( tg)gr1t|glr for some t[2,r1 ] .
 ,(tn|r|l ) } is a set of cardinality nl|r|+1 by the following reasoning : if (ti)=(tj ) with i < j , then (tjti1)=0 , which contradicts 0(t ) . note that a+b = g in view of proposition  2.1(ii ) ; moreover , if |r|>0 , then every element has at least two representations.suppose 0(a+(ti))(a+(tj ) ) for some i < j , i.e. ,  0 has at least two representations , say 0=xig+(ti ) and 0=xjg+(tj ) , as a sum in a+b , where xi , xj[1,n ] .
consequently , since ( from ( 13 ) ) { ( r+1)g,(r+2)g,  ,(l+|r|)g}r+1(glr ) , and since hr+1(glr ) if r is nontrivial ( from ( 14 ) ) , we see from the definition of a that xi , xj[1,r ] , else 0r+1(s ) , contrary to hypothesis .
thus gxiti|s and gxjtj|s are zero - sum subsequences , and so our hypothesis of all zero - sums having length r implies (ti)=(|ti|r)g and (tj)=(|tj|r)g , whence (tjti1)=|tjti1|g .
but now rtjti1 contradicts the maximality of r. therefore we may instead assume that 0 has a unique representation in a+b , in which case r is trivial , as remarked in the previous paragraph.however , in this case a={g,2g,  ,lg } is an arithmetic progression with difference g such that 0a+b = g is a unique expression element . hence it follows that |b{lg,(l1)g,  ,g}|=1 .
,g } , so that ( in view of |b|=|g||a|+1=nl+1 ) ( 19)b={0,g,2g,  ,(nl1)g}{b0}. observe , in view of ( 19 ) and lemma  3.2 , that it now suffices to show |supp(s)|2 to complete the proof .
let tk be the subsequence such that (tk)=b0.note that if we swap the index between gi and gi+1 , for i[1,k1 ] , and use this ordering to define a new b , let us call it b , as above , then b0b and only one element of b differs from b , namely that corresponding to (ti ) .
however , applying the above argument using b instead of b , we see that we again contradict the maximality of r unless b = b ( in view of b0b ) . as b = b
if and only if gi = gi+1 , we conclude that g1=g2==gk . likewise , swapping the index between gi and gi+1 , for i[k+1,nl1 ] , and proceeding as we did for i[1,k1 ] allows us to conclude gk+1=gk+2==gnl .
let g1=ag and gnl = bg , with a , b[2,n1 ] ( since 0,gsupp(t ) ) .
if |t|3 , then we can find an ordering of the gi such that g1=gnl . then using this ordering to define b and repeating the above arguments , we either contradict the maximality of r or show |supp(s)|=2 , in which case the proof is complete as remarked before .
so it only remains to consider the case |t|=2 , as the proof is trivially complete when |t|=1 .
but in this case , l = n2 and a , b[2,n1 ] imply that gna(ag)|s and gnb(bg)|s are both zero - sum subsequences of respective lengths na+1 and nb+1 , whence the uniqueness of r as a zero - sum length implies a = b .
thus supp(s)={g , ag } , completing the proof as remarked before . recalling the well - known fact that a zero - sum free subsequence of length |g|1 must be of the form g|g|1 for a generator gg ( this can be proved in a few lines using the trivial case |b|=k=2 in proposition  2.1(i ) ; see also  [ 9 ,  lemma 5.4.2 ] for a slightly more involved proof ) , we see that the cases r=1 and r = n are trivial .
observe that ( 5)0r1(s)=r1(0r2s),(6)0r+1(s)=n(0nr1s)=(s)nr1(0nr1s ) , where we have used for ( 6 ) the fact that m(t)=(t)|t|m(t ) for tf(g ) , which follows in view of the correspondence between r|t and tr1|t .
we first show that either ( 7)h(s)max{r , nr+1 } ,  or(8)h(s)max{r , nr}ands = gn/2gn/2withord(g)=ord(g)=n  even  , where gg . we do so in two cases .
note that if there are distinct g , gsupp(s ) each with multiplicity at least nr , then ( 9 ) implies that n is even with s = gn/2gn/2 and r = n2 . if ord(g)=ord(g)=n , then ( 8) holds , as desired . on the other hand ,
if ( say ) ord(g)n2 , then ord(g)=r = n2 , in which case the proof is easily concluded using lemma  3.3 .
therefore we may assume that there is at most one term with multiplicity at least nr .
let h = h(nr1(0nr1s ) ) . now assuming h(s)nr , it follows , in view of ( 1 ) , ( 6 ) , and since there is at most one term of multiplicity nr , that h is a proper , nontrivial subgroup .
moreover , in view of v0(0nr1s)=nr1n21|g / h|1 , which follows from ( 9 ) , we see that ( 1 ) implies that all but at most |g / h|2 terms of s are from h. letting t|s be the subsequence of all terms not from h , we see that (s)(t)+h .
thus , since |t||g / h|2n21nr1 ( by ( 9 ) ) , it follows , in view of the definition of h , that (s)nr1(0nr1s ) , in contradiction to ( 6 ) .
therefore we may instead assume that ( 9 ) fails , i.e. , ( 10)r1>n21 .
in this case , we apply theorem  2.2 to r1(0r2s ) . however , assuming h(s)r1 and repeating the above arguments using ( 5 ) instead of ( 6 ) and using ( 10 ) instead of ( 9 ) , we arrive at the same contradiction . therefore we conclude that h(s)r > n2>nr , as claimed .
factor s = glt , where tf(g ) , and let r|t be a maximal length subsequence ( possibly trivial ) such that (r)=|r|g . in view of ( 7 ) , ( 8) and ( 5 ) , it follows that ( 11)vg(s)=l = h(s)max{r , nr}n2|t| and 0(t ) ; in particular , 0(r ) .
then it follows in view of ( 11 ) that r = ord(g ) and that g is the only element from hg in supp(s ) ( else we can find a zero - sum of length distinct from r ) . iteratively applying the definition of d(g / h)|g / h| to h(u1sgord(g ) ) , beginning with u trivial , we find a zero - sum mod h subsequence u|sgord(g ) with |u|n|h||g / h|+1 .
adding on an appropriate number of terms from gord(g ) ( note (gord(g))=h ) yields a zero - sum subsequence u|s with |u|n|h||g / h|+2 . if |h|<n2 , then |u|>|h|=r , a contradiction . on the other hand ,
if |h|=n2 , then we obtain the same contradiction unless |u|=n21 , (u)=g and su1gn/2=g0h .
thus , if there is some g0supp(t)h with g0g0 , then swapping g0 for g0 in u yields a new u|sgord(g ) with (u)h and |u|=n21 but (u)g , whence we obtain the contradiction as before .
therefore , we instead see that all terms outside h in supp(s ) are equal to g0 .
however , since all terms inside h in supp(s ) are equal to g , this shows that |supp(s)|2 .
since |r||t|rln2=ord(g)2 ( the last inequality holds else the proof is complete , while the other inequalities follow from ( 11 ) ) , and since (r)=|r|g , it follows that ( 12)0{g,2g,  ,rg}(gl),(13)0{(r+1)g,(r+2)g,  ,(l+|r|)g}r+1(glr ) . hence l+|r|ord(g)1=n1 and |r|<|t|=nlr .
next , we show that , when r is nontrivial , there is some ( 14)hr+1(glr){g,2g,  ,(l+|r|)g}. thus assume for the moment that r is nontrivial . then , in view of lemma  3.1 and 0(r ) , there is some nontrivial r0|r with (r0){0,g,  ,|r|g}. note (r0)|r|g=(r ) implies |r0|<|r|<|t|r ; thus 1|r0|r2 . if (r0){g,2g,  ,(r|r0|1)g } , then 0r1(glr0 ) , contrary to hypothesis . therefore (r0){(|r|+1)g,(|r|+2)g, 
,(nr+|r0|)g } , whence l+|r|ord(g)1=n1 and gl|s with lrr|r0|+10 show that either (r0)=(nr+|r0|)g=(|r0|r)g or else ( 14 ) holds , as desired .
however , in the former case , factor r = r0r1 and note that (r1)=(r)(r0)=(|r1|+r)g .
now |r|<|t|<|r1|+r|t|1+rn1 , where the last inequality follows from |t|=nl with lr , whence (r1){0,g,  ,|r|g } ( in view of ord(g)=n ) and so |r1|<|r|<|t|r .
thus 1|r1|r2 , and applying the above arguments with r1 instead of r0 , we establish ( 14 ) unless ( |r1|+r)g=(r1)=(|r1|r)g .
however , the latter case implies 2rg=0 , whence r = n2 with n even .
furthermore , by the above work for r0 and r1 , we see that ( 14 ) is established unless ( 15)(r){g,2g,  ,|r|g}{(|r|n2)g } for all nontrivial r|r . applying ( 15 ) to each xsupp(r ) , noting that gsupp(r ) ( in view of r|t ) , and recalling that |r|<|t|r = n2 , we conclude that supp(r){2g,3g,  ,(n21)g}{(n2 + 1)g}. if there are ( n2 + 1)g , xsupp(r ) with x{2g,3g, 
,(n21)g } , then applying ( 15 ) to the sequence x((n2 + 1)g ) yields a contradiction .
,(n21)g}. noting that n2g = rg=rg and (r1)=(|r1|+r)g , we see that { ( |r1|+r+1)g,(|r1|+r+2)g,  ,(n2)g}(r1)+r|r1|2(gl)r2(glr1 ) .
thus , since |r1|+r+1|r|+l+1 , and in view of ( 13 ) and ord(g)=n , we have g{g,(r1)g}r2(glr)r(glr ) . as a result ( recall |r|<|t| ) , ( 17)supp(tr1)={(r1)g}={(n2 + 1)g},(18)(r1)gr2(glr)r(glr ) , else we find a zero - sum of length distinct from r using precisely one term from tr1 ( recall gsupp(t ) in view of the definition of t ) , contrary to hypothesis . by ( 16 ) ,
,(n21)g } , else supp(s)={g , n+22 g } with r = n2 , from which the remainder of the proof is easily deduced .
thus , since r = n2 and r is nontrivial , we see that r3 and ( tg)gr1t|glr for some t[2,r1 ] .
 ,(tn|r|l ) } is a set of cardinality nl|r|+1 by the following reasoning : if (ti)=(tj ) with i < j , then (tjti1)=0 , which contradicts 0(t ) . note that a+b = g in view of proposition  2.1(ii ) ; moreover , if |r|>0 , then every element has at least two representations .
suppose 0(a+(ti))(a+(tj ) ) for some i < j , i.e. ,  0 has at least two representations , say 0=xig+(ti ) and 0=xjg+(tj ) , as a sum in a+b , where xi , xj[1,n ] .
consequently , since ( from ( 13 ) ) { ( r+1)g,(r+2)g,  ,(l+|r|)g}r+1(glr ) , and since hr+1(glr ) if r is nontrivial ( from ( 14 ) ) , we see from the definition of a that xi , xj[1,r ] , else 0r+1(s ) , contrary to hypothesis .
thus gxiti|s and gxjtj|s are zero - sum subsequences , and so our hypothesis of all zero - sums having length r implies (ti)=(|ti|r)g and (tj)=(|tj|r)g , whence (tjti1)=|tjti1|g .
but now rtjti1 contradicts the maximality of r. therefore we may instead assume that 0 has a unique representation in a+b , in which case r is trivial , as remarked in the previous paragraph .
however , in this case a={g,2g,  ,lg } is an arithmetic progression with difference g such that 0a+b = g is a unique expression element .
,g } , so that ( in view of |b|=|g||a|+1=nl+1 ) ( 19)b={0,g,2g,  ,(nl1)g}{b0}. observe , in view of ( 19 ) and lemma  3.2 , that it now suffices to show |supp(s)|2 to complete the proof .
note that if we swap the index between gi and gi+1 , for i[1,k1 ] , and use this ordering to define a new b , let us call it b , as above , then b0b and only one element of b differs from b , namely that corresponding to (ti ) .
however , applying the above argument using b instead of b , we see that we again contradict the maximality of r unless b = b ( in view of b0b ) . as b = b if and only if gi = gi+1 , we conclude that g1=g2==gk .
likewise , swapping the index between gi and gi+1 , for i[k+1,nl1 ] , and proceeding as we did for i[1,k1 ] allows us to conclude gk+1=gk+2==gnl .
let g1=ag and gnl = bg , with a , b[2,n1 ] ( since 0,gsupp(t ) ) . if |t|3 , then we can find an ordering of the gi such that g1=gnl . then using this ordering to define b and repeating the above arguments , we either contradict the maximality of r or show |supp(s)|=2 , in which case the proof is complete as remarked before .
so it only remains to consider the case |t|=2 , as the proof is trivially complete when |t|=1 .
but in this case , l = n2 and a , b[2,n1 ] imply that gna(ag)|s and gnb(bg)|s are both zero - sum subsequences of respective lengths na+1 and nb+1 , whence the uniqueness of r as a zero - sum length implies a = b .
first remark that ord(g)=n holds trivially for |g|=p prime , so step 2 is unnecessary .
next , noting that the case n=2 is trivial , we can assume n3 , and thus that n is odd .
this eliminates the lengthy extra portion of step 3 needed to establish ( 14 ) when r = n2 with n even .
also , the following argument , using the cauchy  davenport theorem instead of the devos  goddyn 
mohar theorem , can be used to establish ( 7).to show ( 7 ) , we proceed in the same two cases .
note that if there are two distinct g , gsupp(s ) with multiplicity at least nr , then this contradicts ( 20 ) in view of n odd , whence we may assume otherwise .
thus , assuming h(s)nr , it is easily seen that we can find nr1 nonempty sets a1,  ,anr1g
such that i=1nr1gaig=0nr1sx1f(g ) , for some xsupp(s ) ( see  [ 1 ,  proposition 2.1 ] ) . applying the cauchy  davenport theorem to a1,  ,anr1 , we find that nr1(0nr1sx1)=g , whence nr1(0nr1s)=g , contradicting ( 6 )
. therefore we may instead assume that ( 20 ) fails , i.e,(21)r > n2 .
in this case , assuming h(s)r1 , we can ( as before ) find r1 nonempty sets a1,  ,ar1g such that i=1r1gaig=0r2sf(g ) . applying the cauchy  davenport theorem to a1,  ,ar1 , we find that r1(0r2s)=g , contradicting ( 5 )
. therefore we conclude , in view of ( 21 ) , that h(s)r > n2>nr , as claimed .
first remark that ord(g)=n holds trivially for |g|=p prime , so step 2 is unnecessary .
next , noting that the case n=2 is trivial , we can assume n3 , and thus that n is odd .
this eliminates the lengthy extra portion of step 3 needed to establish ( 14 ) when r = n2 with n even .
also , the following argument , using the cauchy  davenport theorem instead of the devos  goddyn 
mohar theorem , can be used to establish ( 7 ) . to show ( 7 ) , we proceed in the same two cases .
note that if there are two distinct g , gsupp(s ) with multiplicity at least nr , then this contradicts ( 20 ) in view of n odd , whence we may assume otherwise .
thus , assuming h(s)nr , it is easily seen that we can find nr1 nonempty sets a1,  ,anr1g
such that i=1nr1gaig=0nr1sx1f(g ) , for some xsupp(s ) ( see  [ 1 ,  proposition 2.1 ] ) .
applying the cauchy  davenport theorem to a1,  ,anr1 , we find that nr1(0nr1sx1)=g , whence nr1(0nr1s)=g , contradicting ( 6 )
. therefore we may instead assume that ( 20 ) fails , i.e , ( 21)r > n2 .
in this case , assuming h(s)r1 , we can ( as before ) find r1 nonempty sets a1,  ,ar1g such that i=1r1gaig=0r2sf(g ) . applying the cauchy  davenport theorem to a1,  ,ar1 , we find that r1(0r2s)=g , contradicting ( 5 )
. therefore we conclude , in view of ( 21 ) , that h(s)r > n2>nr , as claimed .
 an old conjecture of graham stated that if n is a prime and s is a sequence of n terms from the cyclic group cn such that all ( nontrivial ) zero - sum subsequences have the same length , then s must contain at most two distinct terms . in 1976 , 
 erds and szemerdi gave a proof of the conjecture for sufficiently large primes n. however , the proof was complicated enough that the details for small primes were never worked out . both in the paper of erds and szemerdi and in a later survey by erds and graham , the complexity of the proof was lamented . 
 recently , a new proof , valid even for non - primes n , was given by gao , hamidoune and wang , using savchev and chen s recently proved structure theorem for zero - sum free sequences of long length in cn . however , as this is a fairly involved result , they did not believe it to be the simple proof sought by erds , graham and szemerdi . in this paper 
 , we give a short proof of the original conjecture that uses only the cauchy  davenport theorem and pigeonhole principle , thus perhaps qualifying as a simple proof . replacing the use of the cauchy  davenport theorem with the devos  goddyn  
 mohar theorem , we obtain an alternate proof , albeit not as simple , of the non - prime case . 
 additionally , our method yields an exhaustive list detailing the precise structure of s and works for an arbitrary finite abelian group , though the only non - cyclic group for which the hypotheses are non - void is c2c2 m . 
Introduction
Notation and preliminaries
Sumsets
Sequences
Subsequence sums
Preliminary results
When the length of a zero-sum is unique