# Factoring finite abelian groups

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Journal of Algebra 275 (2004) Factoring finite abelian groups A.D. Sands Department of Mathematics, Dundee University, Dundee DD1 4HN, Scotland, UK Received 1 April 2003 Communicated by Paul Flavell Abstract It is shown that if in the factorization of a finite cyclic group each factor has either prime power order or order equal to the product of two primes then one factor must be periodic. This is shown to be the best possible result of this type Elsevier Inc. All rights reserved. Keywords: Factorization; Abelian group 1. Introduction In the study of the factorization of finite abelian groups it is usual to ask under what conditions one of the factors must be periodic. With the exception of elementary 2-groups this leads to formulae [8] giving all factorizations where these conditions hold. The conditions may involve the structure of the group and the number of factors. Alternately they may involve numerical conditions on the orders of the factors. The best known result of this second type is the famous Theorem of Rédei [5] which asserts that if each factor has prime order then one factor must be a subgroup. It should be noted that a periodic factor of prime order is a subgroup. For cyclic groups, the more general result holds [8] that if each factor has prime power order then one factor must be periodic. From examples given by de Bruijn [2], it follows that for cyclic groups this last result does not extend to factors of order p 2 q or pqr,wherep, q, r are distinct primes, even if the remaining factors have prime order. No example is known to exclude factors of order pq. We shall see in Theorem 1 that the conclusion does hold for factors of this type and that address: /\$ see front matter 2004 Elsevier Inc. All rights reserved. doi: /j.jalgebra

2 A.D. Sands / Journal of Algebra 275 (2004) we may also admit factors of prime power order. We shall then show that this result along with a previous result from [8] gives a complete solution in the case of cyclic groups. This result can also be used to greatly shorten the proofs of previous results in [1,6]. In the non-cyclic case, we show in Theorem 2 that if the p-primary component of a group G is cyclic and if all factors except perhaps one have order a power of the prime p then some factor must be periodic. This generalizes results in [6,8,9] and also provides shorter proofs for the results in [6,9]. Finally we list some open problems of this type for the non-cyclic case. 2. Preliminaries Throughout the paper the word group will be used to mean finite abelian group. The additive notation is used. The sum of subsets A 1,A 2,...,A r of a group G is the subset of all elements of the form r i=1 a i,wherea i A i for each i. The sum is said to be direct if each such element has a unique expression in this form. If the direct sum of subsets equals G, we call this a factorization of the group G. The notation A + g is used to denote the set of all elements {a + g: a A}. In a factorization, each subset A i may be replaced by A i + g i for any elements g i G. Hence we may and do assume that 0 A i for each factor A i. The order of the subset A of G is denoted by A. In order to avoid trivial cases, we assume that A i > 1 for each factor A i of G. A subset A of a group G is said to be periodic if there exists a non-zero element g of G such that A + g = A. ThesetH of these periods, together with 0, forms a subgroup of G. Clearly A is a union of cosets of H. Equivalently, there exists a subset D such that A = H + D, whered is non-periodic. If G = A 1 + A 2 + +A r is a factorization in which A 1 = H + D 1 then we obtain a factorization of the quotient group G/H. If, again, one factor is periodic, this process may be continued. In [8] formulae are given to obtain all such factorizations whenever this process continues at each stage. This subject arose because of the solution by Hajós [4] of one of the classical problems of Minkowski. Whether or not a factor must be periodic continues to be of interest because of the existence of these formulae. A group has been called good if in every factorization involving two factors one factor must be periodic. The group is called k-good if this holds true for factorizations involving k factors. A factorization is said to be bad if none of the factors is periodic. The cyclic group of order n is denoted by Z(n). The subgroup generated by a subset A of a group G is denoted by A.IfA and B are non-periodic subsets and if the sum A +B is direct then it has been shown [7] that A + B is non-periodic. If H is a proper non-zero subgroup of a group G then there exists a non-periodic complete set of coset representatives C for G modulo H except in the case where H has index 2 in an elementary 2-group G [7]. If we have a factorization of a proper non-zero subgroup H of a group G into r nonperiodic factors and if C is a non-periodic complete set of coset representatives then G has bad factorizations with r + 1 factors and also with r factors. H = A 1 + +A r leads to G = A 1 + +(A r + C) and the result follows from above by regarding A r + C as either two factors or one single factor.

3 542 A.D. Sands / Journal of Algebra 275 (2004) When dealing with these problems, Rédei [5] introduced the use of group characters. If χ is a character of a group G and A is a subset of G then χ(a)is defined to be the complex number a A χ(a).ifg = A 1 + +A r is a factorization of G then it is easily seen that χ(g)= χ(a 1 ) χ(a r ).Ifχ is not the unity character then χ(g)= 0 and so there exists i such that χ(a i ) = 0. We have equivalent methods for cyclic groups using cyclotomic polynomials. Let g be a generator of Z(n). IfA is a subset of order r then there exist integers d i,1 i r, 0 d i <n, such that A ={d i g:1 i r}. We associate with A the polynomial A(x) = ri=1 x d i, which has degree less than n. Sincec d(mod n) if and only if x c x d (mod (x n 1)), it follows that from a factorization G = A 1 + +A r we obtain that A 1 (x) A r (x) G(x) = 1 + x + +x n 1 (mod (x n 1)). IfF d (x) denotes the dth cyclotomic polynomial then if d divides n and d>1 it divides both G(x) and x n 1. Since F d (x) is irreducible over the field Q of rationals there exists i such that it divides A i (x). There is, in fact, an old result of Kronecker, who shows that if d and m are relatively prime then F d (x) is irreducible over the field of mth roots of unity. The special case where d = p e is a prime power is most useful. If we reduce the exponents in A i (x) modulo p e we obtain a polynomial B i (x) with degree less than p e andsuchthata i (x) B i (x) (mod (x pe 1)). Hence F p e(x) = 1 + x pe x (p 1)pe 1 divides B i (x). Since the quotient is a polynomial of degree less than p e 1 this leads to considerable information about B i (x).in particular, the coefficients of x s and x s+pe 1 are equal for 0 s<(p 1)p e 1. As has been mentioned, for cyclic groups the two methods are essentially equivalent. If χ is a character of Z(n) with χ(g) = ρ, whereρ is a dth primitive root of unity, then χ(a i ) = A i (ρ). NowA i (ρ) = 0 if and only if F d (x) divides A i (x). So the methods are equivalent but deductions can often be obtained more easily by using cyclotomic polynomials. In this paper we make use of Proposition 3 from [10]. We make one useful deduction here. For a subset A of a group G,wedefinenA to be the family of elements {na: a A} for each integer n. Let a group G be a direct sum of subgroups H, K of relatively prime orders. For g G, we use the notation g = g H + g K,whereg H H, g K K. Wedefine the H -component of A by A H ={a H : a A}. Lemma. Let a group G be a direct sum of subgroups H,K of relatively prime orders. Let G = A + B be a factorization of G such that A divides H.ThenG = A H + B is also a factorization. Proof. Let H =m, K =n. Sincem and n are relatively prime, there exists k such that kn 1 (mod m). Since A divides m, it follows that A and kn are relatively prime. By [10, Proposition 3] it follows that kna + B = G is a factorization. Since kna = a H, it follows that A H + B = G is a factorization. We should note that this implies that the elements {a H : a A} are distinct.

4 A.D. Sands / Journal of Algebra 275 (2004) Cyclic groups Before proceeding to the general case of the result given in the introduction for cyclic groups, we observe that the result is essentially known for cyclic groups with only two distinct prime factors. If Z(p e q f ) = A 1 + +A r is a factorization then F p e qf (x) divides A i (x) for some i. If A i is a prime power then it is already known [8] that A i is periodic. If A i =pq then a standard application of [3, Theorem 2] yields A i (1) = pq = pp(1) + qq(1) where P(x),Q(x) are polynomials with non-negative integer coefficients. Thus either P(1) = q,q(1) = 0orP(1) = 0,Q(1) = p. Q(1) = 0 implies Q(x) = 0andsothat p e 1 q f g is a period of A i,whereg is a generator of G. Similarly P(1) = 0 implies that p e q f 1 g is a period of A i. So, in this case, if the given conditions are satisfied one factor must be periodic. Theorem 1. Let G be a cyclic group and let there be a factorization of G in which each factor has either prime power order or order equal to the product of two primes. Then one of the factors is periodic. Proof. Let G have order n and generator g. Letχ be a character of G such that χ(g) is an nth primitive root of unity. It follows from [8] that if χ(a i ) = 0where A i is a prime power then A i is periodic. So we may assume that G = A 1 + +A u + +A k,where χ(a i ) = 0 if and only if i u and that, for these values of i, A i =p i q i where p i, q i are distinct primes. Let B i be the (G pi + G qi )-component of A i. By the lemma we may replace A i by B i to obtain the factorization G = B 1 + +B u + A u+1 + +A k.from above for some value of i we must have χ(b i ) = 0. We may assume that χ(b 1 ) = 0 and, for convenience, that B 1 =pq. Let n = p e q f m,wherem is not divisible by p or by q.leta, b, c of orders p e, q f, m, respectively, be a generating set for G. As already noted, since B 1 is a subset of Z(p e q f ), χ(b 1 ) = 0 implies that B 1 is periodic. Since B 1 =pq, it has a period of order p or of order q. Without loss of generality, we may assume that p e 1 a is a period of B 1.Let H = p e 1 a. Then there is a subset D of order q such that B 1 = H + D. By the lemma, the elements of B 1 andsoalsoofd are distinct. Let D = q j=1 (r j a + s j b). From the form of H we may assume that 0 r j <p e 1. Then from B 1 = H + D it follows that p 1 A 1 = i=0 j=1 q (( ip e 1 ) + r j a + sj b + e(i,j)c ) where 0 s j <q f,0 e(i,j) < m.

5 544 A.D. Sands / Journal of Algebra 275 (2004) Since χ(g) has order n, it follows that χ(a)= α, χ(b)= β, χ(c)= γ have orders p e, q f, m, respectively. Then χ(a 1 ) = 0 implies that p 1 i=0 j=1 q α ip(e 1) +r j β s j γ e(i,j) = 0. It follows that the polynomial obtained by replacing α by x is divisible by F p e(x). Thus the coefficients of x r, x r+pe 1,...,x r+(p 1)pe 1 in this polynomial are equal. Hence β s j γ e(0,j) = = β s j γ e(p 1,j). r j =r r j =r As before, it follows that F q f (x) divides, for each i and i, the polynomial x s j γ e(i,j) x s j γ e(i,j). r j =r r j =r So the coefficients in this polynomial of x s, x s+qf 1,...,x s+(q 1)qf 1 are all equal. Now the pairs (r j,s j ) are distinct. Thus, for a given pair (r, s + tq f 1 ), there is either a unique j with (r j,s j ) equal to this pair or else no such j exists at all. So the coefficient in the above polynomial of x s+tqf 1 is either of the form γ e(i,j) γ e(i,j) or else is 0 as no such j exists. Suppose for a given r that both situations arise. Then all coefficients are 0 for this value of r. Thus γ e(i,j) = γ e(i,j) for the v values of t for which such coefficients arise, where 1 v<q. Since the occurrence of (r j,s j ) does not involve i, this occurs for all pairs i and i. So for this value of r, vp terms are involved. For any other value, say r,theremust be fewer than pq terms involved and so the same result arises. Hence e(i,j) = e(i,j) for all i, i and j. Therefore A 1 is periodic with period p e 1 a. So we may assume that r, s exist such that all q values of (r, s + tq e 1 ),0 t<q,do occur. This gives all pq terms in A 1 and, since 0 A 1, it follows that B 1 is the subgroup of order pq.sointhiscasewehave A 1 = p 1 q 1 ( ip e 1 a + jq f 1 b + e(i,j)c ). i=0 j=0 Then, as above, we obtain that for all i, i, j, j, γ e(i,j) γ e(i,j) = γ e(i,j ) γ e(i,j ). These are complex numbers of modulus 1 and from an Argand diagram it is easy to see that they must be equal in pairs. So three cases can arise as follows:

6 A.D. Sands / Journal of Algebra 275 (2004) (C1) e(i,j) = e(i,j ) and e(i,j)= e(i,j ); (C2) e(i,j) = e(i,j) and e(i,j ) = e(i,j ); (C3) e(i,j) = e(i,j ) + m/2 and e(i,j)= e(i,j ) + m/2. The third case, in which γ e(i,j) = γ e(i,j ), can only arise if m is even. We recall that 0 e(i,j) < m and all sums are being taken modulo m. Firstly let us assume that only cases C1 and C2 occur. If, for a given i, e(i,j) = e(i,j ) for all j, j then it follows that for all i, j, j we have e(i,j)= e(i,j ). Hence q f 1 b is a period of A 1. So we may suppose that for each i there exists s,s such that e(i,s) e(i,s ). Then C2 holds for this pair s, s and for all i.soe(i,s) = e(i,s).byc2,e(i,j ) = e(i,j ) for all i, j. Hence e(i,j ) = e(r,j )(= e(i,j )) for all i, r, j. It follows that A 1 is periodic with period p e 1 a. Now we may suppose that r, r and s, s exist such that neither C1 nor C2 is satisfied. Then m must be even and e(r,s) = e(r,s ) + m/2,e(r,s)= e(r,s ) + m/2. Let γ e(r,s) = ρ,γ e(r,s) = σ.thenasγ m/2 = 1wehavethatγ e(r,s ) = ρ, γ e(r,s ) = σ. Since neither C1 nor C2 is satisfied it follows that ρ σ, ρ σ. Since m is even, it follows that pq is odd and so by [10, Proposition 3] that we may replace A 1 by 2A 1.Ifχ(2A 1 ) = 0 then, as above using 2a, 2b as generators, we obtain that γ 2e(i,j) γ 2e(i,j) = γ 2e(i,j ) γ 2e(i,j ). For r, r and s, s, this implies that ρ 2 σ 2 = σ 2 ρ 2.Thisgivesρ = σ or ρ = σ,which is false. We may now proceed by induction on u.ifu = 1, we would have the contradiction of a factorization G = 2A 1 + A 2 + +A k in which χ(a) is not zero for any factor A. Thus, in this case, C1 or C2 must hold always and from the above we have that A 1 is periodic. For u>1, this new factorization has only u 1 factors, A say, with χ(a) = 0. The required result follows by induction on u. This completes the proof. This result leads to shorter proofs of certain earlier results, since the proof here does not depend on these results. In [6] lengthy proofs are given that the groups Z(p 2 qr) and Z(pqrs), wherep, q, r, s are distinct primes, are good. The lengthy cases are for factors of orders pq,pr in the first group and for factors of order pq, rs in the second. These now follow immediately from Theorem 1. Corrádi and Szabó [1] gave a lengthy proof where one factor has order pq and the other factors have prime order. They did this to answer a question raised in [8]. Again this result follows immediately from Theorem 1. We should note that the C3 case can arise in the above proof and that χ(a 1 ) = χ(b 1 ) = 0 need not imply that A 1 is periodic. Consider the following example which uses the notations as above. Let p, q be distinct odd primes and let m be an even integer greater than 2 and not divisible by p or q.letz(pqm) be generated by elements a,b,c of orders p, q, m, respectively. Let I ={i: 0 i p 1} be a disjoint union of non-empty subsets

7 546 A.D. Sands / Journal of Algebra 275 (2004) I 1, I 2 with 0 I 1. Similarly, let J ={j: 0 j q 1} be a disjoint union of non-empty subsets J 1, J 2 with 0 J 1.Definee(i,j) as follows: 0 if (i, j) I 1 J 1, m/2 if (i, j) I e(i,j) = 2 J 2, 1 if (i, j) I 2 J 1, 1 + m/2 if (i, j) I 1 J 2. Let A 1 = p 1 q 1 i=0 j=0 (ia + jb+ e(i,j)c). ThenB 1 is the subgroup a,b = Z(pq) and so χ(b 1 ) = 0. Now χ(a 1 ) = α i β j α ( ) i β j + γ α i β j α i β j i I 1 j J 1 i I 2 j J 2 i I 2 j J 1 i I 1 j J 2 = α i β j α ( i β j + γ α i β j α i β ). j i I j J 1 i I 2 j J i I j J 1 i I 1 j J Hence χ(a 1 ) = 0, as i I αi = 0 = j J βj. A 1 is not periodic since A 1 =pq and neither a nor b is a period. A 1 does occur in such factorizations of G. IfH = c then since B 1 = a,b it follows that G = A 1 + H and H may be factored into factors of suitable orders. Of course, one of these factors of H will be periodic. In [8, Theorem 2] it is shown that if G is cyclic and every factor in a factorization except perhaps one has order a power of the same prime p then one factor must be periodic. In the introduction we claimed that we now have the best possible result of this type. This follows from examples given by de Bruijn [2] and then by the method of extending examples of bad factorizations already mentioned. Let G contain a proper subgroup H which is the direct sum of two subgroups H 1,H 2 of composite order. Then H 1, H 2 contain non-zero subgroups K 1, K 2 and non-periodic sets of coset representatives A 1, A 2, respectively, with H 1 = K 1 + A 1, H 2 = K 2 + A 2. De Bruijn shows how to construct a third non-periodic subset A 3 such that G = A 1 + A 2 + A 3.We should note that A 1 + A 2 is also non-periodic, since A 1 H 1 and H 1 + A 2 is direct. In order that H should be cyclic, we need the condition that H 1, H 2 should be relatively prime. Thus this holds also for A 1, A 2. We may choose A 1, A 2 to be of distinct prime orders. Since A 3 = G/H K 1 K 2, we may choose G so that A 3 is of order p 2 q or pqr where these are distinct primes. Now, as mentioned previously, we can extend bad factorizations of subgroups of prime index to bad factorizations of the group and either include a new factor of prime order or increase the order of a given factor by a prime multiple. Thus, for a given set of orders of factors, provided that one factor has order divisible by p 2 q or pqr, where these are distinct primes, and the orders of the remaining factors taken together involve at least two distinct primes, bad factorizations with this set of orders of factors exist. As claimed, this leaves only the sets of orders contained in [8, Theorem 2] and in Theorem 1.

8 A.D. Sands / Journal of Algebra 275 (2004) Non-cyclic groups We have mentioned above the result from [8] that if all factors but one in the factorization of a cyclic group have orders equal to powers of the same prime then one factor is periodic. There is also a result [9] that states that if the p-component of a group is cyclic then in any factorization involving only two factors if one factor has order a power of the prime p then again one of the two factors must be periodic. We show here that these two results may be generalized by allowing all factors but one to be of order a power of the same prime p, but with the retention of the condition that the p-component is cyclic. Theorem 2. Let p be a prime and let G be a group such that the p-component G p is cyclic. If G = A 1 + +A r + B is a factorization of G such that each factor A i has order equal to a power of p then one of the factors is periodic. Proof. For each i let B i be the G p -component of A i. By the lemma, A i may be replaced by B i and the elements of B i are distinct. Let H be the subgroup of G which is the complement of G p. Since the subsets B i are contained in G p it follows that, for each h H, G p + h = B B r + (B (G p + h)). This leads to the factorization G p = B 1 + +B r + ((B (G p + h)) h). SinceG p is a cyclic group of prime power order it follows that one of the factors must be periodic. Suppose first that no factor B i is periodic. Let G p have order p e and generator a.then p e 1 a is a period of (B (G p + h)) h and so of B (G p + h) for each h H. Since, as h varies over H, B is the union of these sets it follows that p e 1 a is a period of B. We may now suppose that one of the other subsets, say B 1, is periodic. Then p e 1 a is a period. We should note that this implies that the subsets B 2,...,B r are not periodic since the sum with B 1 being direct implies that p e 1 a cannot also be a period of one of these sets. From the results given in [8], it follows that if χ is a character such that χ(a) has order p e then χ(b i ) 0for2 i r. Now let us consider the factorization G = A 1 + B 2 + +B r + B. For any character χ such that χ(a) has order p e, we have that either χ(a 1 ) = 0orthatχ(B) = 0. Let χ j, j J, be the set of all such characters such that χ j (A 1 ) = 0. Let the kernel of χ j be K j, i.e. K j ={g G: χ j (g) = 1}. Let the intersection of all these subgroups K j be K. Let F = p e 1 a. We note that the statement that χ(a) has order p e is equivalent to χ(p e 1 a) 1 and so to the statement that χ(f)= 0. Let us suppose that K = 0. Let χ j (A 1 ) = 0. If h H and ra + h A 1 then ra B 1 and so only one such r can exist. χ j (A 1 ) = 0 implies that α r χ j (h) = 0, where the summation is taken over all ra + h A 1.ThenF p e(x) divides x r χ j (h). Sincep e 1 a is a period of B 1, it follows that if sa + h 1 is in A 1 with h 1 H then there exists h 2 H such that (s + p e 1 )a + h 2 A 1. From above it follows that χ j (h 1 ) = χ j (h 2 ) andsothat h 1 h 2 K j. This is true for all such j. Hence h 1 h 2 is in K. It follows that h 1 = h 2 andsothatp e 1 a is a period of A 1. Finally we have the case in which K 0. Let χ be a character such that χ(f) = χ(k) = 0. χ(k) = 0 implies that χ(a 1 ) 0. Thus χ(b) = 0. By [11, Theorem 2] it follows that there exist subsets U, V of G such that B is the disjoint union of the direct sums F + U and K + V.NowthesumB 1 + B is direct and p e 1 a is a period of B 1.It

9 548 A.D. Sands / Journal of Algebra 275 (2004) follows that U is the empty set and so that B = K + V. Thus every non-zero element of K is a period of B. This completes the proof. This provides a better proof for the result in [9] which considers the case with just one factor of order a power of p. The result in [8] is a routine extension of the result for one such factor in [6, Theorem 2], but we now have an alternative proof for this result. In the cyclic case, we were able to determine which sets of orders of factors ensured that one factor is periodic. This problem is far from solved in the non-cyclic case. We close the paper by listing some unsolved problems. De Bruijn [2] has given an example in which there is a bad factorization with factors of order m and mk, wherem 4, k>1, and G contains a subgroup of type Z(m)+ Z(m). By choosing m to be a prime greater than 3 and k to be any prime and extending this factorization as before, we are left only with Rédei s Theorem, i.e. the case where every factor has prime order. In the case of 2-groups examples of bad factorizations with factors of order 8 and 4 and with factors of order 8, 2 and 2 are known from [2]. No bad factorization is known with factors of orders 2 and 4 only. So we have the following question. Problem 1. If a group G is factorized as a sum of factors each of which has prime order or order 4 must one of the factors be periodic? In the case of 3-groups, there are examples of bad factorizations in [2] with factors of orders 3k and 3m for any k, m greater than 1 provided that G has a subgroup isomorphic to Z(3) + Z(3). This leaves unsolved the case where G has a single factor of order 9. For other groups, it leaves unsolved also the case where G has a single factor of order 6. So we have the following question. Problem 2. If a group G is factorized as a sum of factors all of which have prime order apart from a single factor of order 6 or 9 must one of the factors be periodic? There is a reference above to an example where the factors have orders m and mk.ifwe restrict attention to cases where the orders of factors are relatively prime, we need to return to the example used for cyclic groups. Now the restriction that subgroups whose sums are direct must have relatively prime orders can be dropped. This allows the possibility of the factor A 1 having order p 3 as well as the previous possible orders p 2 q and pqr. One of the factors in the m and mk case must have non-prime order. So we have the following question. Problem 3. If a group G is factorized as a sum of factors each of which has prime order or order the product of two, not necessarily distinct, primes and if the order of each factor which is not prime is relatively prime to the product of the orders of the remaining factors must then one factor be periodic? It may be possible to combine the sets of orders in Problems 1 and 2. It may also be possible to weaken some of the conditions in Problem 3 for cases involving the primes two

10 A.D. Sands / Journal of Algebra 275 (2004) and three. Probably, it is better not to speculate about the detailed results which may hold here until some significant progress has been made with the general case. References [1] K. Corrádi, S. Szabó, Solution to a problem of A.D. Sands, Comm. Algebra 23 (1995) [2] N.G. de Bruijn, On the factorization of finite abelian groups, Indag. Math. 15 (1953) [3] N.G. de Bruijn, On the factorization of cyclic groups, Indag. Math. 15 (1953) [4] G. Hajós, Über einfache und mehrfache Bedeckung des n-dimensionalen Raumes mit einem Wurfelgitter, Math. Z. 47 (1942) [5] L. Rédei, Die neue Theorie der endlichen Abelschen Gruppen und Verallgemeinerung des Hauptsatzes von Hajós, Acta Math. Hungar. 16 (1965) [6] A.D. Sands, On the factorization of finite abelian groups, Acta Math. Hungar. 8 (1957) [7] A.D. Sands, Factorization of abelian groups II, Quart. J. Math. 13 (1962) [8] A.D. Sands, Factorization of cyclic groups, in: Proc. Coll. Abelian Groups (Tihany), 1964, pp [9] A.D. Sands, On the factorization of finite abelian groups III, Acta Math. Hungar. 25 (1974) [10] A.D. Sands, Replacement of factors by subgroups in the factorization of abelian groups, Bull. London Math. Soc. 32 (2000) [11] A.D. Sands, S. Szabó, Factorization of periodic subsets, Acta Math. Hungar. 57 (1991)

### FACTORING CERTAIN INFINITE ABELIAN GROUPS BY DISTORTED CYCLIC SUBSETS

International Electronic Journal of Algebra Volume 6 (2009) 95-106 FACTORING CERTAIN INFINITE ABELIAN GROUPS BY DISTORTED CYCLIC SUBSETS Sándor Szabó Received: 11 November 2008; Revised: 13 March 2009

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### 3.1 The Definition and Some Basic Properties. We identify the natural class of integral domains in which unique factorization of ideals is possible.

Chapter 3 Dedekind Domains 3.1 The Definition and Some Basic Properties We identify the natural class of integral domains in which unique factorization of ideals is possible. 3.1.1 Definition A Dedekind

### A Hajós type result on factoring finite abelian groups by subsets II

Comment.Math.Univ.Carolin. 51,1(2010) 1 8 1 A Hajós type result on factoring finite abelian groups by subsets II Keresztély Corrádi, Sándor Szabó Abstract. It is proved that if a finite abelian group is

### ALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS. 1. Ideals in Commutative Rings In this section all groups and rings will be commutative.

ALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS PETE L. CLARK 1. Ideals in Commutative Rings In this section all groups and rings will be commutative. 1.1. Basic definitions and examples. Let R be a (commutative!)

### The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### Introduction to Finite Fields (cont.)

Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number

### SUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by

SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples

### 7. Some irreducible polynomials

7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of

### Cyclotomic Extensions

Chapter 7 Cyclotomic Extensions A cyclotomic extension Q(ζ n ) of the rationals is formed by adjoining a primitive n th root of unity ζ n. In this chapter, we will find an integral basis and calculate

### ON GENERALIZED RELATIVE COMMUTATIVITY DEGREE OF A FINITE GROUP. A. K. Das and R. K. Nath

International Electronic Journal of Algebra Volume 7 (2010) 140-151 ON GENERALIZED RELATIVE COMMUTATIVITY DEGREE OF A FINITE GROUP A. K. Das and R. K. Nath Received: 12 October 2009; Revised: 15 December

### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### some algebra prelim solutions

some algebra prelim solutions David Morawski August 19, 2012 Problem (Spring 2008, #5). Show that f(x) = x p x + a is irreducible over F p whenever a F p is not zero. Proof. First, note that f(x) has no

### FACTORING ABELIAN GROUPS OF. Department of Mathematics, University of Bahrain, P. O. Box

Mathematica Pannonica 7/2 (1996), 197 { 207 FACTORING ABELIAN GROUPS OF ORDER p 4 Sandor Szabo Department of Mathematics, University of Bahrain, P. O. Box 32038 Isa Town, State of Bahrain Khalid Amin Department

### 6.2 Permutations continued

6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of

### 3 Congruence arithmetic

3 Congruence arithmetic 3.1 Congruence mod n As we said before, one of the most basic tasks in number theory is to factor a number a. How do we do this? We start with smaller numbers and see if they divide

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express

### 3 1. Note that all cubes solve it; therefore, there are no more

Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)

Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)

### Finite fields: further properties

Chapter 4 Finite fields: further properties 8 Roots of unity in finite fields In this section, we will generalize the concept of roots of unity (well-known for complex numbers) to the finite field setting,

### Galois Theory III. 3.1. Splitting fields.

Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where

### ADDITIVE GROUPS OF RINGS WITH IDENTITY

ADDITIVE GROUPS OF RINGS WITH IDENTITY SIMION BREAZ AND GRIGORE CĂLUGĂREANU Abstract. A ring with identity exists on a torsion Abelian group exactly when the group is bounded. The additive groups of torsion-free

### Chapter 13: Basic ring theory

Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring

### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing

### MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

### GROUPS SUBGROUPS. Definition 1: An operation on a set G is a function : G G G.

Definition 1: GROUPS An operation on a set G is a function : G G G. Definition 2: A group is a set G which is equipped with an operation and a special element e G, called the identity, such that (i) the

### 11 Ideals. 11.1 Revisiting Z

11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will

### Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN. Part II: Group Theory

Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN Part II: Group Theory No rights reserved. Any part of this work can be reproduced or transmitted in any form or by any means. Version: 1.1 Release: Jan 2013

### 26 Ideals and Quotient Rings

Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 26 Ideals and Quotient Rings In this section we develop some theory of rings that parallels the theory of groups discussed

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### A REMARK ON ALMOST MOORE DIGRAPHS OF DEGREE THREE. 1. Introduction and Preliminaries

Acta Math. Univ. Comenianae Vol. LXVI, 2(1997), pp. 285 291 285 A REMARK ON ALMOST MOORE DIGRAPHS OF DEGREE THREE E. T. BASKORO, M. MILLER and J. ŠIRÁŇ Abstract. It is well known that Moore digraphs do

### Finite Sets. Theorem 5.1. Two non-empty finite sets have the same cardinality if and only if they are equivalent.

MATH 337 Cardinality Dr. Neal, WKU We now shall prove that the rational numbers are a countable set while R is uncountable. This result shows that there are two different magnitudes of infinity. But we

### A simple criterion on degree sequences of graphs

Discrete Applied Mathematics 156 (2008) 3513 3517 Contents lists available at ScienceDirect Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam Note A simple criterion on degree

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### Factorization in Polynomial Rings

Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important

### A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number

Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

### Prime Numbers and Irreducible Polynomials

Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.

3. QUADRATIC CONGRUENCES 3.1. Quadratics Over a Finite Field We re all familiar with the quadratic equation in the context of real or complex numbers. The formula for the solutions to ax + bx + c = 0 (where

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

### DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS

DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS ASHER M. KACH, KAREN LANGE, AND REED SOLOMON Abstract. We construct two computable presentations of computable torsion-free abelian groups, one of isomorphism

### COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS

Bull Austral Math Soc 77 (2008), 31 36 doi: 101017/S0004972708000038 COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS IGOR V EROVENKO and B SURY (Received 12 April 2007) Abstract We compute

### The fundamental group of the Hawaiian earring is not free (International Journal of Algebra and Computation Vol. 2, No. 1 (1992), 33 37) Bart de Smit

The fundamental group of the Hawaiian earring is not free Bart de Smit The fundamental group of the Hawaiian earring is not free (International Journal of Algebra and Computation Vol. 2, No. 1 (1992),

### Introduction to finite fields

Introduction to finite fields Topics in Finite Fields (Fall 2013) Rutgers University Swastik Kopparty Last modified: Monday 16 th September, 2013 Welcome to the course on finite fields! This is aimed at

### Sign changes of Hecke eigenvalues of Siegel cusp forms of degree 2

Sign changes of Hecke eigenvalues of Siegel cusp forms of degree 2 Ameya Pitale, Ralf Schmidt 2 Abstract Let µ(n), n > 0, be the sequence of Hecke eigenvalues of a cuspidal Siegel eigenform F of degree

### The finite field with 2 elements The simplest finite field is

The finite field with 2 elements The simplest finite field is GF (2) = F 2 = {0, 1} = Z/2 It has addition and multiplication + and defined to be 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 0 0 = 0 0 1 = 0

### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

### 4.1 Modules, Homomorphisms, and Exact Sequences

Chapter 4 Modules We always assume that R is a ring with unity 1 R. 4.1 Modules, Homomorphisms, and Exact Sequences A fundamental example of groups is the symmetric group S Ω on a set Ω. By Cayley s Theorem,

### ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS

ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS John A. Beachy Northern Illinois University 2014 ii J.A.Beachy This is a supplement to Abstract Algebra, Third Edition by John A. Beachy and William D. Blair

### 1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain

Notes on real-closed fields These notes develop the algebraic background needed to understand the model theory of real-closed fields. To understand these notes, a standard graduate course in algebra is

### CONTRIBUTIONS TO ZERO SUM PROBLEMS

CONTRIBUTIONS TO ZERO SUM PROBLEMS S. D. ADHIKARI, Y. G. CHEN, J. B. FRIEDLANDER, S. V. KONYAGIN AND F. PAPPALARDI Abstract. A prototype of zero sum theorems, the well known theorem of Erdős, Ginzburg

### COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS

COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS IGOR V. EROVENKO AND B. SURY ABSTRACT. We compute commutativity degrees of wreath products A B of finite abelian groups A and B. When B

### Abelian groups/1. 1 Definition

Abelian groups 1 Definition An Abelian group is a set A with a binary operation satisfying the following conditions: (A1) For all a,b,c A, we have a (b c) = (a b) c (the associative law). (A2) There is

### Galois representations with open image

Galois representations with open image Ralph Greenberg University of Washington Seattle, Washington, USA May 7th, 2011 Introduction This talk will be about representations of the absolute Galois group

### Lecture 1 (Review of High School Math: Functions and Models) Introduction: Numbers and their properties

Lecture 1 (Review of High School Math: Functions and Models) Introduction: Numbers and their properties Addition: (1) (Associative law) If a, b, and c are any numbers, then ( ) ( ) (2) (Existence of an

### Integral Domains. As always in this course, a ring R is understood to be a commutative ring with unity.

Integral Domains As always in this course, a ring R is understood to be a commutative ring with unity. 1 First definitions and properties Definition 1.1. Let R be a ring. A divisor of zero or zero divisor

### JUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson

JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials

### 4. FIRST STEPS IN THE THEORY 4.1. A

4. FIRST STEPS IN THE THEORY 4.1. A Catalogue of All Groups: The Impossible Dream The fundamental problem of group theory is to systematically explore the landscape and to chart what lies out there. We

### Group Theory. Contents

Group Theory Contents Chapter 1: Review... 2 Chapter 2: Permutation Groups and Group Actions... 3 Orbits and Transitivity... 6 Specific Actions The Right regular and coset actions... 8 The Conjugation

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### A characterization of trace zero symmetric nonnegative 5x5 matrices

A characterization of trace zero symmetric nonnegative 5x5 matrices Oren Spector June 1, 009 Abstract The problem of determining necessary and sufficient conditions for a set of real numbers to be the

### 2. Let H and K be subgroups of a group G. Show that H K G if and only if H K or K H.

Math 307 Abstract Algebra Sample final examination questions with solutions 1. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30 and 40, Determine H. Solution. Since gcd(18,

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. In Memory of Robert Barrington Leigh. Saturday, March 5, 2016

THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION In Memory of Robert Barrington Leigh Saturday, March 5, 2016 Time: 3 1 2 hours No aids or calculators permitted. The grading is designed

### G = G 0 > G 1 > > G k = {e}

Proposition 49. 1. A group G is nilpotent if and only if G appears as an element of its upper central series. 2. If G is nilpotent, then the upper central series and the lower central series have the same

### Sets and set operations

CS 441 Discrete Mathematics for CS Lecture 7 Sets and set operations Milos Hauskrecht milos@cs.pitt.edu 5329 Sennott Square asic discrete structures Discrete math = study of the discrete structures used

### CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

### CHAPTER 5: MODULAR ARITHMETIC

CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called

### Appendix A. Appendix. A.1 Algebra. Fields and Rings

Appendix A Appendix A.1 Algebra Algebra is the foundation of algebraic geometry; here we collect some of the basic algebra on which we rely. We develop some algebraic background that is needed in the text.

FINITE FIELDS KEITH CONRAD This handout discusses finite fields: how to construct them, properties of elements in a finite field, and relations between different finite fields. We write Z/(p) and F p interchangeably

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### Math 210A: Algebra, Homework 1

Math 210A: Algebra, Homework 1 Ian Coley October 9, 2013 Problem 1. Let a 1, a 2,..., a n be elements of a group G. Define the product of the a i s by induction: a 1 a 2 a n = (a 1 a 2 a n 1 )a n. (a)

### r + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.

Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in

### Chapter 5: Rational Functions, Zeros of Polynomials, Inverse Functions, and Exponential Functions Quiz 5 Exam 4

Chapter 5: Rational Functions, Zeros of Polynomials, Inverse Functions, and QUIZ AND TEST INFORMATION: The material in this chapter is on Quiz 5 and Exam 4. You should complete at least one attempt of

### Math 3000 Running Glossary

Math 3000 Running Glossary Last Updated on: July 15, 2014 The definition of items marked with a must be known precisely. Chapter 1: 1. A set: A collection of objects called elements. 2. The empty set (

### 4. CLASSES OF RINGS 4.1. Classes of Rings class operator A-closed Example 1: product Example 2:

4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets

### Topics in Number Theory

Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall

### Computer Algebra for Computer Engineers

p.1/14 Computer Algebra for Computer Engineers Preliminaries Priyank Kalla Department of Electrical and Computer Engineering University of Utah, Salt Lake City p.2/14 Notation R: Real Numbers Q: Fractions

### Section 3.3 Equivalence Relations

1 Section 3.3 Purpose of Section To introduce the concept of an equivalence relation and show how it subdivides or partitions a set into distinct categories. Introduction Classifying objects and placing

### Applications of Methods of Proof

CHAPTER 4 Applications of Methods of Proof 1. Set Operations 1.1. Set Operations. The set-theoretic operations, intersection, union, and complementation, defined in Chapter 1.1 Introduction to Sets are

### GROUPS WITH TWO EXTREME CHARACTER DEGREES AND THEIR NORMAL SUBGROUPS

GROUPS WITH TWO EXTREME CHARACTER DEGREES AND THEIR NORMAL SUBGROUPS GUSTAVO A. FERNÁNDEZ-ALCOBER AND ALEXANDER MORETÓ Abstract. We study the finite groups G for which the set cd(g) of irreducible complex

### Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients

DOI: 10.2478/auom-2014-0007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### Algebra 2. Rings and fields. Finite fields. A.M. Cohen, H. Cuypers, H. Sterk. Algebra Interactive

2 Rings and fields A.M. Cohen, H. Cuypers, H. Sterk A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 1 / 20 For p a prime number and f an irreducible polynomial of degree n in (Z/pZ)[X ], the quotient

### calculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,

Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials

### CONSEQUENCES OF THE SYLOW THEOREMS

CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important

### On the Exponential Diophantine Equation (12m 2 + 1) x + (13m 2 1) y = (5m) z

International Journal of Algebra, Vol. 9, 2015, no. 6, 261-272 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ija.2015.5529 On the Exponential Diophantine Equation (12m 2 + 1) x + (13m 2 1) y