Spectra of Semidirect Products of Cyclic Groups

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1 Spectra of Semidirect Products of Cyclic Groups Nathan Fox 1 University of Minnesota-Twin Cities Abstract The spectrum of a graph is the set of eigenvalues of its adjacency matrix A group, together with a multiset of elements of the group, gives a Cayley graph, and a semidirect product provides a method of producing new groups This paper compares the spectra of cyclic groups to those of their semidirect products, when the products exist It was found that many of the interesting identities that result can be described through number theory, field theory, and representation theory The main result of this paper gives a formula that can be used to find the spectrum of semidirect products of cyclic groups 1 Introduction Given a graph, its spectrum is defined as the set of eigenvalues of its adjacency matrix The adjacency matrix encodes all of the information about a graph in a compact form A group, together with a multiset of elements, gives a Cayley graph Semidirect products are an interesting way of producing small, non-abelian groups The semidirect products that we will consider are built from cyclic groups We will examine the spectra of cyclic groups as well as those of their semidirect products, when the products exist Our main technical result Theorem 41) is a general formula which factors the characteristic polynomial of the adjacency matrix of the Cayley graph of a semidirect product of cyclic groups as a product of characteristic polynomials of some simple matrices, as follows: χ A C Z/n k Z/m, S))) = χ Ω ia C m ) b x a y b S 1 This reasearch was carried out at Canisius College with funding from the National Science Foundation The author would like to thank Dr Terrence Bisson for his assistance 1

3 eigenvalues, or characteristic polynomials, more quickly than building a large adjacency matrix and taking a determinant Specifically, we will look at a type of directed graph that is derived from a group: a Cayley graph Definition 22 Given a group G and a multiset S such that s S for all s S, the Cayley graph with generators S is a directed graph with one vertex corresponding to each group element, and for each pair of elements g 1, g 2 G there is an edge from g 1 to g 2 for each element s S such that g 1 s = g 2 Denote this graph as C G, S) Note that S generates the graph, but not necessarily the group The Cayley graph will be connected if and only if S generates the group It is more useful, though, to consider arbitrary multisets of elements of the group More specifically, we shall examine Cayley graphs of specific groups that can be built from less complicated pieces: semidirect products of cyclic groups 21 Semidirect Products of Groups In general, spectra of Cayley graphs can be quite complicated The spectra of finite abelian groups are known, but comparatively little is known about the spectra of even the smallest non-abelian groups The least complicated non-abelian groups can be built from cyclic groups using a semidirect product Definition 23 Given two groups G and H and a group homomorphism ϕ : H Aut G), the Semidirect Product of G and H with respect to ϕ, denoted G ϕ H or, simply, G H) is a new group with set G H and multiplication operation g 1, h 1 ) g 2, h 2 ) = g 1 ϕ h 1 ) g 2, h 1 h 2 ) In practice, Definition 23 can be complicated to use Luckily, when G and H are both cyclic, there is a nice presentation For this paper, we will use multiplicative notation for cyclic groups, where Z/n is generated by an element x such that x n = e Proposition 21 Given cyclic groups Z/n and Z/m, a semidirect product Z/n Z/m between them corresponds to a choice of integer k such that k m 1 mod n) The semidirect product group is given by Z/n Z/m = x, y x n = e, y m = e, yxy 1 = x k, and will be denoted Z/n k Z/m 3

4 A proof of this proposition can be found in [2] The idea is that k gives a group homomorphism from Z/m to Aut Z/n) When constructing an adjacency matrix for a Cayley graph of a semidirect product of cyclic groups, we will always assume that the vertices are ordered such that the first row and column of the matrix correspond to the identity, the next n 1 rows and columns correspond to powers of x in increasing order, and then each block of n rows and columns corresponds to the powers of y in ascending order and within each block, the powers of x take the same order) 3 Cayley Graphs and Representations Representation theory is the study of embedding groups as subgroups of GL N F) for some integer N and some field F The embedding map is a homomorphism ψ : G GL N F), and we say that the representation is faithful if ψ is injective In this paper, we are mainly concerned with representations of groups of order n embedded in GL n C), or, more specifically, GL n Q [ω]) for some root of unity ω Note that all of the fields that we are concerned with have characteristic zero Given a group G and an element g G, let A g = A C G, g})), the adjacency matrix of the Cayley graph with one generator Additionally, given a group G and a multiset S of elements of G, let A S = A C G, S)) Theorem 31 Given a group G and an element g G, consider the set Γ = A g g G} and the map ψ : G Γ given by ψ g) = A g Then, ψ gives a faithful representation for G in GL G Q) Proof Consider X = C G, G) Each matrix in Γ determines a subgraph of X Consider two matrices A g1, A g2 Γ The matrix A g1 gives the number of paths in X of length one from a group element h 1 to a group element h 2 following only paths corresponding to multiplication by g 1 Similarly, A g1 gives the number of paths in X of length one from a group element h 1 to a group element h 2 following only paths corresponding to multiplication by g 2 Thus, A g1 A g2 gives the number of paths in X of length two from a group element h 1 to a group element h 2 following only paths corresponding to a multiplication by g 1 followed by a multiplication by g 2 This is equivalent to following only paths of length one corresponding to multiplication by g 1 g 2 4

5 Thus, A g1 A g2 = A g1 g 2 Therefore, it is clear that Γ is a faithful representation for G with injection ψ Based on this theorem, we can define a group representation based on the adjacency matrices of Cayley graphs Definition 31 The adjacency representation of a group G is the representation given by ψ This is called the regular representation in the literature [5] The next theorem will allow for simple computation of adjacency matrices when multiple generators are used Theorem 32 Given a group G and a multiset S of elements of G, A S = s S A s Proof The proof is by induction on S For the base case, S = 1 This means that S = s} for some s G It is obviously true that A S = A s = A r r s} Now, as an inductive hypothesis, assume that if S < h, then A S = s S A s Let S = h S = T s} for some s G and some multiset T Clearly, every edge in C G, T ) is present in C G, S) because T S All of the additional edges come from C G, s}) Thus, A S = A T + A s By the inductive hypothesis, A S = A s + A t = A s t T s S as required 5

6 31 Preliminary Notation and Results The following definitions and propositions will be extremely important for the remainder of the paper They will be used, beginning in the next section, to describe representations of semidirect products Definition 32 Let C h be the h h matrix with entries 1, when j i 1 mod h) c ij = This matrix is denoted C h because it is a circulant matrix [?davis] C could also denote "cyclic", as C h is the adjacency matrix for the Cayley graph of Z/h with S = x} where x h = e) Proposition 31 The matrix C h ) d is given by 1, when j i d mod h) c ij = Proof The first part of the proof is by induction on d For the base case, let d = 1 The definition of C h completes the proof of this case, as C h ) 1 = C h Now, let b > 1 Assume that C h ) b 1 is given by the specified formula C h ) b = C h ) b 1 C h, and entries that are one in this matrix are those where the i th row of C h ) b 1 has a one in the k th column and the k th row of C h has a one in the j th column All other entries are zero This occurs when k i b 1 mod h) j k 1 mod h) Solving the first equation for k yields k b 1+i mod h) Substituting this into the other equation gives j b + 1 i 1 mod h), or j i b mod h), as required Now, let d < 1 Note that, by the previous steps, C h ) h = I Thus, C h ) d = C h ) h) m Ch ) d = C h ) mh+d Since h > 0, there must exist an integer m such that the quantity mh + d > 0, and this quantity will be congruent to d modulo h Thus, the proposition holds for all integers d 6

7 Definition 33 Suppose that m, n, and k satisfy m k 1 mod n) For given h, let let Ω h be the m m matrix with entries ω hk i 1, when i = j Ω ij = where ω = e 2πi n is a primitive n th root of unity In general, the primitive h th root of unity e 2πi h will be denoted ω h, but sometimes the subscript will be omitted if the meaning seems clear) This matrix is denoted Ω h because it contains roots of unity, which are denoted by ω Proposition 32 Powers of the matrix Ω h are given by Ω h ) a = Ω ha Proof Ω h ) a is an m m matrix with entries ω hak i 1, when i = j Ω ij = This is clearly equal to Ω ha 32 Representations of Semidirect Products of Cyclic Groups When it is known that G is a semidirect product of cyclic groups, another representation in GL G C) can be found This new representation will have a form such that computation of characteristic polynomials, and hence, eigenvalues, is easier than in the adjacency representation First, however, it is useful to examine exactly what form the matrices in the adjacency representation take For the following theorem, recall that A g is the adjacency matrix of a Cayley graph with one generator Let x be a generator for Z/n and y a generator for Z/m Theorem 33 For a semidirect product Z/n k Z/m, A x is an m m block matrix with n n matrix entries given by Cn ) x ij = ki 1, when i = j 7

8 and A y is an m m block matrix with n n matrix entries given by I, when j i 1 mod n) y ij = Proof Consider the group element g = x a y b gx = x a y b x = x a+kb y b, so A x is in the required form gy = x a y b+1, so A y is in the required form Now, we can find a representation such that computation is easier Theorem 34 Let X be an n n block matrix with m m matrix entries given by Ωi, when i = j x ij = and let Y be an n n block matrix with m m matrix entries given by y ij = Cm, when i = j The matrices X and Y generate a faithful representation of Z/n k Z/m with injection ϕ such that ϕ x a y b) = X a Y b Proof We will show that X n = I, Y m = I, and Y XY 1 = X k, thereby precisely showing that ϕ produces a representation Showing that no smaller power of X or Y is trivial will show that ϕ is injective X a is an n n block matrix with m m matrix entries given by Ωi ) a = Ω x ij = ia, when i = j Since Ω h ) n = I for all h by Proposition 32, X n = I Also, note that Ω 1 is a diagonal matrix containing a primitive n th root of unity in the upper left corner Thus, no smaller power of this matrix, and, hence, no smaller power of X, can be the identity Y b is an n n block matrix with m m matrix entries given by Cm ) b, when i = j y ij = Based on Proposition 31, Y m = I, and no smaller power of Y is trivial as no smaller power of C m is trivial) 8

9 Y 1 is an n n block matrix with m m matrix entries given by Cm ) 1, when i = j y ij = Thus, Y XY 1 is an n n block matrix with m m matrix entries given by Cm Ω a ij = i C m ) 1, when i = j To show that this equals X k it suffices to show that C m Ω h C m ) 1 = Ω hk C m Ω h is an m m matrix with entries ω hk j 1, when j i 1 mod m) a ij = Multiplying this by C m ) 1 as given by Proposition 31) yields a ij = ω hk j, when i = j This precisely equals Ω hk, thereby completing the proof Definition 34 The natural representation of Z/n k Z/m is the representation given by ϕ We say that two representations M and N of a group G are isomorphic if the matrices M g and N g corresponding to a group element g are similar that is, one can be obtained from the other via a change of basis) and for all group elements, that change of basis is the same This allows the isomorphism to be applied to linear combinations of representation matrices as well: Let P be the change of basis matrix to move from one representation to an isomorphic one, let A and B be matrices in the first representation, and let A and B be their corresponding matrices in the second representation so A = P AP 1 and B = P BP 1 ) Then, for scalars a and b, P aa + bb) P 1 = P aap 1 + P bbp 1 = ap AP 1 + bp BP 1 = aa + bb, which corresponds to aa + bb by the same isomorphism Theorem 35 The adjacency representation and the natural representation of Z/n k Z/m are isomorphic group representations 9

10 Proof It is well-known that two representations over a field of characteristic zero are isomorphic if the traces of corresponding matrices are the same [5] I will show that tr ) A x a y = tr X a Y b) in all cases Note that x a y b = e if and b only if X a Y b = I Clearly tr A e ) = tr I) = mn This proves the identity case Now, consider representations of the element x a y b, where x a y b e The adjacency representation of this element is A x a y b C Z/n k Z/m, x a y b}) has no self-loops, so all of the diagonal entries of A x a yb are zero Thus, tr ) A x a y = 0 The natural representation of this element is X a Y b Clearly, b X a has matrix entries given by Ωi ) a = Ω x ij = ia, when i = j and Y b has matrix entries given by Cm ) b, when i = j y ij = Thus, X a Y b has matrix entries given by Ωi ) a C xy ij = m ) b = Ω ia C m ) b, when i = j Clearly, if b 0 mod m), this matrix has zeroes down the diagonal, and, therefore, has trace 0 Consider the case where b = 0 I will show that tr X a ) = 0 if a 0 mod n)) m 1 tr X a ) = tr Ω ia ) = ω iakj Now, assume that iak j = r for some value of r Clearly, r = sa for some s Thus, we have ik j = s, or i = k j s Thus, for each value of s, there will be m terms in the double sum that equal sa This means that the double sum equals ) ω m ω i n 1 = m = 0 ω 1 as required Therefore, these two group representations are isomorphic 10 j=0

11 Now that we know that these representations are isomorphic, we can convert between the two representations at will Any group-theoretic statement that is true with the adjacency representation is also true with the natural representation In particular, corresponding matrices will have the same characteristic polynomial This fact will be quite important in the proof of the main theorem, in the next section 4 Characteristic Polynomials of Semidirect Products of Cyclic Groups The following is our main result about characteristic polynomials of semidirect products of cyclic groups It can be applied in numerous specific cases to yield information about the spectra of Cayley graphs of semidirect products of cyclic groups Sometimes, it can even lead to explicit formulas for the eigenvalues Theorem 41 The characteristic polynomial of the semidirect product of two cyclic groups is given by the following: χ A C Z/n k Z/m, S))) = χ Ω ia C m ) b x a y b S Proof Let G = Z/n k Z/m By Theorem 32, ) ) χ A C G, S))) = χ A C G, s})) = χ A s s S Since s G, it can be written uniquely as x a y b for some 0 a < n and some 0 b < m Thus, the formula becomes χ A x a y = χ X a Y b, b x a y b S x a y b S by Theorem 35 Then, by Definition 34 and Theorem 35), it becomes χ Ω i ) a C m ) b = χ Ω ia C m ) b x a y b S 11 x a y b S s S

12 by Proposition 32, as required The rest of this section and all of the next section will show a variety of ways in which this formula can be applied The easiest examples allow for direct computation of eigenvalues, whereas other applications only allow for computation of a characteristic polynomial 41 Spectra of Finite Abelian Groups An immediate application of Theorem 41 is to the calculation of spectra of Cayley graphs of finite abelian groups For example, the following theorem about the spectra of cyclic groups was proved at the 2006 REU project at Canisius [4] Proposition 41 The eigenvalues of C Z/n, S) are given by } λ λ = s S ω xs, x Z, 1 x n Proof Clearly, Z/n = Z/n Z/1 = Z/n 1 Z/1 That means that Ω h = I 1 ω h Thus, by Theorem 41, ) ) χ A C Z/n, S))) = χ Ω ia = χ I 1 ω ia x a S ) = χ I 1 ω ia which is clearly the required formula x a S x a S = λ ) ω ia x a S A much more general result can also be proved regarding spectra of finite abelian groups Theorem 42 Let x 1,, x h be generators for cyclic groups Z/n 1,, Z/n h The eigenvalues of the Cayley graph of this product group with generators S has eigenvalues } h ω j ia i n i, 0 j b < n b, λ λ = s S i=1 12

13 where each s in the sum is written as for some sequence of values a i s = Proof According to [5], a representation with matrices of dimension h b=1 h n h i=1 can be built as a tensor product from the natural representations of the cyclic groups, and it will clearly be isomorphic to the analog built as a tensor product of the adjacency representations All of the matrices in this representation will be diagonal, so the eigenvalues of their linear combinations will be the linear combinations of their entries These are precisely the eigenvalues specified by the formula Corollary 41 Let x be a generator for Z/n, and let y be a generator for Z/m The eigenvalues of C Z/n Z/m, x, y}) are λ λ = ω i n + ω j m, 0 i < n, 0 j < m } x a i i h i=1 n h Proof Apply Theorem 42 with h = 2, n 1 = n, n 2 = m, and S = x, y} 42 Examples of Spectra of Semidirect Products of Cyclic Groups In addition to confirming known results about abelian groups, Theorem 41 can also be used to investigate spectra of non-abelian semidirect products The least complicated such groups are dihedral groups Definition 41 The dihedral group of order 2n, denoted D 2n, is Z/n 1 Z/2 Theorem 41 leads to a general form for the characteristic polynomials of Cayley graphs of dihedral groups with arbitrary generators 13

14 Theorem 43 χ A C D 2n, S))) = λ 2 λ ω ia + ω ia) + ω ia ω ib x a S x a S x b S x a S x a y S x b y S Proof Applying Theorem 41 yields χ A C D 2n, S))) = χ Ω i + ) C 2 x a S x a y S Ω i + C 2 = ω ia x a S ω ia x a y S x a y S ω ia x a y S ω ia x a S ω ia ω ib This is a 2 2 matrix with complex entries It has characteristic polynomial λ ) ω ia λ ) ) ) ω ia ω ia ω ia x a S x a S x a y S x a y S = λ 2 λ ω ia + ω ia) + ω ia ω ib ω ia ω ib x a S x a S x b S x a y S x b y S Substituting this into the original formula yields λ 2 λ ω ia + ω ia) + ω ia ω ib x a S x a S as required x b S x a y S x b y S ω ia ω ib An application of Theorem 43 leads to the following theorem, which was shown at the 2006 REU at Canisius [1] Corollary 42 χ A C D 2n, x, y}))) = λ n χ A C Z/n, ±1}))) 14

15 Proof Applying Theorem 43 yields χ A C D 2n, x, y}))) = λ 2 λ ω i + ω i) + ω i ω i ω i ω i)) = λ λ ω i + ω i)) by Proposition 41, as required = λ n λ ω i + ω i)) = λ n χ A C Z/n, ±1}))) Another relatively well-behaved type of semidirect product is that formed between two cyclic groups of odd prime order Theorem 44 is another application of Theorem 41 Theorem 44 Let p 1 and p 2 be odd primes such that p 1 divides p 2 1 It is well known that this condition is necessary and sufficient for a nontrivial semidirect product to exist [2]) Let k give a nontrivial semidirect product Then, χ A C Z/p 2 k Z/p 1, x, y}))) = for some polynomial q λ) p 2 1 p1 1 j=0 ) λ ω ikj) 1 = λ 1) p 1 + 1) q λ) p 1 Theorem 44 will require two lemmas to be proven The first lemma gives the beginnings of a form for the characteristic polynomials of these groups This lemma will be presented in a more general form than required to prove the theorem, as it holds for any semidirect product Z/n k Z/m For the remainder of this paper, given m, n, and k satifying m k 1 mod n), let Z i = Ω i + C m, where Ω i and C m are both m m Lemma 41 For a semidirect product Z/n k Z/m, for every value of h, χ Z h ) = m 1 j=0 λ ω hkj) 1 15

16 Proof Of course χ Z h ) = det λi Z h ) This matrix has binomials on the main diagonal, ones on the superdiagonal and in the lower left corner, and zeroes elsewhere By Leibniz s formula, det A = mn sgn σ) a i,σi) σ S mn i=1 In λi Z h, choosing a nonzero element in the first row amounts to choosing a nonzero element in either the last row if the binomial on the main diagonal is chosen) or in the second row if the -1 on the superdiagonal is chosen) It is clear that this process propagates so that the only permutations choosing only nonzero elements are the one that selects the diagonal entries the identity permutation, which is even) and the one that selects the superdiagonal elements and the lower left element, which is a cycle of length m, and, hence, has sign 1) m+1 Since this term without the sign is the product of -1 m times, it is true that as required χ Z h ) = det λi Z h ) = m 1 j=0 λ ω hkj) 1 Let p 2 be an odd prime In order to state the next lemma, we need the following Definition 42 Let a, b Z Let be the relation on Z/p 2 if a = k d b for some d Z given by a b Proposition 42 is an equivalence relation Proof I will show that satisfies the axioms of an equivalence relation Reflexivity: a = ak 0, so a a Symmetry: Let a b This means that there exists an integer d such that ak d = b Note that bk d = a, where k d is defined as the inverse of k modulo p 2 raised to the d power This means that b a Transitivity: Let a b and b c This means that there exist integers d 1 and d 2 such that ak d 1 = b and bk d 2 = c Note that ak d 1 k d 2 = ak d 1+d 2 = c This means that a c Therefore, is an equivalence relation 16

17 The second lemma establishes equality of characteristic polynomials of blocks within the partitions specified by Definition 42 Lemma 42 When working with a semidirect product of cyclic groups of odd prime order so n = p 2 and m = p 1, if a b, then χ Z a ) = χ Z b )) Proof First, I will show that all of the equivalence classes given by have the same size Consider an arbitrary element h Z/p 2 The partition of Z/p 2 containing h is m m = hk d for some d } Recall that k p 1 1 mod p 2 ) Since p 1 is prime, no smaller power of k can be one Thus, the size of the partition containing h must be p 1 Since h was arbitrary, all partitions must have size p 1 and, hence, there are p 2 1 p 1 of them) Each matrix Z contains p 1 roots of unity on the diagonal The powers on ωp b 2 are clearly one partition of Z/p 2 Thus, if a b, the element in the upper left corner in Z b will appear somewhere on the diagonal of Z a Since this element s exponent is in both partitions, they must be the same partition Thus, the diagonal elements are the same; they are just in a different order This results in the same characteristic polynomial, as polynomial multiplication over Z [ω p2 ] is commutative Now that all of the necessary machinery is in place, we can prove Theorem 44 Proof In this proof, the Z matrices are determined as they were in Lemma 42 By Theorem 41, χ A C Z/p 2 k Z/p 1, x, y}))) = χ Z i ) = χ Z 0 ) χ Z i ) By Lemma 41, χ Z 0 ) = λ 1) p 1 + 1), as needed By Lemmas 42 and 42, χ Z i ) i=1 is a perfect p th 1 power, as there are equivalences of characteristic polynomials over partitions of size p 1 This concludes the proof Note that, in general, the entries in the matrices Z n are complex numbers, and, hence, their characteristic polynomials have complex coefficients Theorem 44 implies that the product of all of these characteristic polynomials is a polynomial with integer coefficients 17 i=1

18 5 Additional Results Many of the calculations in this paper work in a cyclotomic field extension of the rational numbers When performing calculations in this field, an identity arises that provides a connection between various mathematical entities This result provides a connection between roots of unity, matrices with integer coefficients, and determinants of block matrices In general, finding the determinant of a block matrix is difficult, but in this specific case we get a simpler answer Let M = A x + A y, the m m block matrices described in Theorem 33 Theorem 51 χ M) = m 1 j=0 ) λ ω ikj) 1 = det m 1 j=0 ) λi C n ) kj) I The proof of Theorem 51 will require the following lemma Lemma 51 χ M) = det m 1 j=0 ) λi C n ) kj) I Proof The formula χ M) = det λi M) is derived by solving the equation Mv = λv so that the values of λ that are roots of the characteristic polynomial are the eigenvalues of M Since M is a block matrix, I will solve for the eigenvalues in a different way Like before, start with Mv = λv Now, since M is an m m block matrix, express v as an m 1 block matrix Let the i th block in v be denoted v i For v 1 example, if m = 3, then v = v 2 This yields the following system of m v 3 equations: C n ) kj v j + v jmod m)+1 = λv j for j Z/ m + 1) 0}) These equations can be rewritten into the form λi C n ) kj) v j = v jmod m)+1 for j Z/ m + 1) 0}) Starting from any one of these equations, substitutions can be done in a cyclic manner until the same vector appears on 18

19 both sides of a single equation Keeping in mind that all matrices of the form λi C n ) kj commute with each other, the following equation is the result of such a substitution into the last equation: m 1 λi C n ) kj)) v 1 = v 1 j=0 Rearranging yields m 1 j=0 ) λi C n ) kj) I v 1 = 0 This yields the desired result that the eigenvalues of M are given by the roots of m 1 ) det λi C n ) kj) I j=0 so this must be an expression for χ M) Now, Theorem 51 can be proved Proof The left side, by Lemma 41, equals χ M) The right side, by Lemma 51, equals χ M) 51 Future Directions It would be useful to find a statement analogous to Theorem 41 for semidirect products of abelian groups in general, as opposed to only for semidirect products of cyclic groups Such a tool could be used to analyze groups such as A 4 = Z/2 Z/2) Z/3 Perhaps an even more general result could be found that would yield information about spectra for any semidirect product, or, more optimistically, for any finite group References [1] T Coon, Combinatorics of the Figure Equation on Directed Graphs, Rose-Hulman Institute of Technology: Undergraduate Math Journal ) [2] D Dummit and R Foote, Abstract Algebra: Second Edition, Prentice Hall, Upper Saddle River, NJ,

20 [3] C Godsil and G Royle, Algebraic Graph Theory, Springer, New York, 2001 [4] J Lazenby, Circulant Graphs and Their Spectra, Senior Thesis, Portland, OR, May 2008 [5] J Serre, Linear Representations of Finite Groups, Springer, New York,

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