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Dedekind Group: a Conclusion of Commutative Group © 2018 JETIR October 2018, Volume 5, Issue 10 www.jetir.org (ISSN-2349-5162) DEDEKIND GROUP: A CONCLUSION OF COMMUTATIVE GROUP Manjeel Kumar Assistant Professor Department of Mathematics D.A.V.College, Hoshiarpur, Punjab. Abstract: The main aim of this paper to analyze completely a special kind of group known as Dedekind group i.e a group in which every subgroup is normal. Here we discuss the commutative condition for a group to become Dedekind group. Keywords: Commutative group, Cyclic group, Homomorphism, Invariant group, Quotient group. 1. Introduction In group theory the concept of Dedekind group was introduced by Julius Wilhelm Richared Dedekind, a German Scientists, who made important contributions to abstract algebra particularly ring theory[10].As every abelian group has every subgroup is invariant so that such type of group is always Dedekind group but on the other hand if we talk about the non abelian group then for that we have an important result given in the book by instance, Hall [p. 190, Theorem 12. 5.4][8] known as structure theorem.It states that “A non abelian group is a Dedekind group if and only if it is the direct product of a quaternion group, an elementary abelian 2-group and an abelian group of odd order”[1]. Preliminaries 2.1. 퐴 푔푟표푢푝 < 퐺,∗> 푠 푠푎푑 푡표 푏푒 푐표푚푚푢푡푎푡푣푒 푔푟표푢푝 푓푓 푎 ∗ 푏 = 푏 ∗ 푎 ∀ 푎, 푏 ∈ 퐺.[2],[3],[4],[5]. 2.2 A group < 퐺, . > is said to be cyclic if ∃ an element ξ ∈ G s.t every element of G can be expressed as power of ξ. We denote it as G =< ξ >.[4],[5]. 2.3 A subgroup H of a group G is said to be invariant or self-conjugate subgroup if Ha = aH ⩝ a ∈ H. It is also known as normal subgroup. Moreover for any a ∈ G ∃ a integer m s. t a = ξm 2.4 A group < 퐺, . > is said to be Dedekind group if every subgroup of G is invariant or normal in G.[1] 2.5 퐿푒푡 < 퐺,∗> 푎푛푑 < 퐺′, ∎ > 푏푒 푎푛푦 푡푤표 푔푟표푢푝푠 . 푇ℎ푒푛 푎 푓푢푛푐푡표푛 휑: 퐺 → 퐺′ 푠 푠푎푑 푡표 푏푒 퐻표푚표푚표푟푝ℎ푠푚 푓 푤푒 ℎ푎푣푒 휑(푎 ∗ 푏) = 휑(푎)∎휑(푏)푤ℎ푒푟푒 푎, 푏 ∈ 퐺. 푁표푡푒: ()퐼푓 퐺 = 퐺′푤푡ℎ 푠푎푚푒 푐표푚푝표푠푡표푛 푡ℎ푒푛 푡 푠 푐푎푙푙푒푑 푒푛푑표푚표푟푝ℎ푠푚 [5],[6],[7] () 푀표푒표푣푒푟 푛 푎푑푑푡표푛 푡표 퐺 = 퐺′, 푓 휑: 퐺 → 퐺 푠 표푛푒 표푛푒 푡ℎ푒푛 푡 푏푒푐표푚푒푠 푎푢푡표푚표푟푝ℎ푠푚 [5], [7], [9]. 3. Theoretical Results 3.1 Theorem: Every subgroup of a Dedekind group is Dedekind. Proof: Let H is subgroup of Dedekind group < 퐺, . > and K be any subgroup of H.Therfore H is normal in G and we have K ⊆ H ⊆ G.Therefore K is normal in G. Hence H is also Dedekind. Note : Clearly {e} and G are trivial or improper Dedekind subgroup of Dedekind group G. Any other Dedekind subgroup other than {e} and G is known as non-trivial or proper Dedekind subgroup of G. 3.2 Theorem : Prove that a group in which every element is of order 2, is a Dedekind group. Proof: Let α, βϵ G ⇒ α2 = e and β2 = e where e is identity element of G. α = α−1, β = β−1.Similarly αβ = (αβ)−1 = β−1α−1 = βα ⇒ G is an commutative group and every subgroup of G is invariant in G, hence Dedekind. 3.3 Theorem: Prove that every group of order less than or equal to 4 is a Dedekind group. Solution: Let G be group of order less than or equal to 4. We have the following cases Case I. If O(G) = 1 i. e. G = {e} clearly G is Dedekind. Case II. If O(G) = 2 i. e. G = {e, α}where α ≠ e.Again G is abelian hence Dedekind. Case III. If O(G) = 3 i. e G = {e, α, β}where α ≠ β ≠ e As αβ ϵ G ⇒ αβ = α or αβ = β or αβ = e but possible case is αβ = e similarly we get βα = e ∴ αβ = βα Hence G is commutative group and every subgroup of G is invariant in G, therefore G is Dedekind group. JETIR1810957 Journal of Emerging Technologies and Innovative Research (JETIR) www.jetir.org 382 © 2018 JETIR October 2018, Volume 5, Issue 10 www.jetir.org (ISSN-2349-5162) Case IV If O(G) = 4 i. e G = {e, α, β, γ} where α ≠ β ≠ γ ≠ e Suppose G is non abelian say αβ ≠ βα Moreover αβ ≠ e because if αβ = e ⇒ βα = e ∴ αβ ≠ βα ≠ α ≠ β ≠ e ∴ G contains at least five elements. Which is not possible . ∴ G is commutative, hence dedekind . From all the cases we can conclude that every group of order less than or equal to four is dedekind 3.4 Theorem : Prove that a finite semi-group with cross cancellation law is a Dedekind group. Proof: For any x, y ∈ G, where G is group satisfied cross cancellation law. By associative law x(yx) = (xy)x ⇒ yx = xy ∴ G is abelian, Hence dedekind. Theorem: If a finite group G of order n contains an element of order n then G is dedekind group. Proof:Since G contains an element of order n ∴ G be a cyclic group and every cyclic group is abelian. Hence G is dedekind group Converse of above result need not be true. Consider the Quatrinonian group Q8 = {±1, ±i, ±j, ±k}. Q8 is a Hamiltonoian group i. e. non abelian dedekind group but does not contains any element of order 8. Theorem: If every cyclic subgroup of G is normal in G then G is a Dedekind group. Proof: Let H be any subgroup of G and a ∈ G Consider a cyclic subgroup K =< 푎 > 표푓 퐺. ∴ K is normal in G ⇒ g−1ag ∈ K ⊆ H , Hence H is normal in G ∴ G is dedekind group. 3.5 Theorem: Let H1, H2, H3,………..Hn be proper invaiant subgroups of group < 퐺, . > 푤ℎ푒푟푒 Hi ∩ Hj = {e} ∀i ≠ j s. t. G = H1 ∪ H2 ∪ H3 ∪ … … … ∪ Hn. Then prove that G is dedekind group. Proof: Let a, bϵG be any elements. If aϵHi and b ∈ Hj where i ≠ j then clearly ab = ba. if a, b ∈ Hi Since Hi is proper subgroup. therefore for any g ∈ G ∃ a subgroup Hk s. t. g ∈ Hk where k ≠ i. ⇒ ga = ag, bg = gb and g(ab) = (ab)g as ab ∈ Hi Now, ga ∉ Hi as g ∉ Hi ∴ (ga)b = b(ga) Now, (ab)g = g(ab) = (ga)b = b(ga) = b(ag) = (ba)g ⇒ ab = ba using cancellation law From both the cases G is abelian , hence dedekind group. G 3.6 Theorem: If G be cyclic group such that ⁄Z(G) is cyclic , then show that G is a dedekind group. G G ( ) Proof: Since ⁄Z(G) is cyclic say ⁄Z(G) =< 푍 G ξ > ( ) ( ) G For any x, y ∈ G ⇒ Z G x, Z G y ∈ ⁄Z(G) ⇒ Z(G)x = (Z(G)ξ)m, Z(G)y = (Z(G)ξ)n where m, n ∈ Z ⇒ Z(G)x = Z(G)ξm, Z(G)y = Z(G)ξn ⇒ xξ−m, yξ−n ∈ Z(G) ∴ x = gξm, y = g′ξn where g, g′ ∈ G Now xy = gξmg′ξn = gg′ξm+n = yx ∀ x, y ∈ G Hence G is commutative group and every subgroup of G is invariant in G, therefore G is Dedekind group. 3.7(a) Theorem: Prove that if the mapping ∅: G → G defined by ∅(x) = x2 , where G be a group, is a homomorphism then G is dedekind group. Proof: For all x, y ∈ G, we have ∅(xy) = ∅(x)∅(y) JETIR1810957 Journal of Emerging Technologies and Innovative Research (JETIR) www.jetir.org 383 © 2018 JETIR October 2018, Volume 5, Issue 10 www.jetir.org (ISSN-2349-5162) ⇒ (xy)2 = x2y2 ⇒ (xy)(xy) = (xx)(yy) ⇒ x(yx)y = x(xy)y ⇒ yx = xy ∀x, y ∈ G Hence G is commutative group and every subgroup of G is invariant in G, therefore G is Dedekind group. 3.7(b) Theorem: Prove that if the mapping ∅: G → G defined by ∅(x) = x−1 , where G be agroup, is homomorphism then G is dedekind group. Proof: For all x, y ∈ G, we have ∅(xy) = ∅(x)∅(y) −1 −1 −1 ⇒ (xy) = x y −1 −1 −1 −1 ⇒ y x = x y −1 −1 −1 ⇒ xy x = y −1 −1 ⇒ yxy x = e ⇒ yx = xy ∀ x, y ∈ G ∴ G is a commutative group, hence dedekind 3 3.7(c )Theorem: Prove that if the mapping ∅: G → G defined by ∅(x) = x , where G be agroup, is a automorphism then G is dedekind group. Proof:For g1, g2 ∈ G −3 3 −3 3 3 −3 3 3 −1 −1 We have ∅(g1 g2g1) = (g1 g2g1) = g1 g2g1 = ∅(g1 )∅(g2)∅(g1) = ∅(g1 g2g1) −3 3 −1 ⇒ g1 g2g1 = g1 g2g1 2 2 ⇒ g2g1 = g1g2 … … … … … . (i) Now ∅(g1g2) = ∅(g1)∅(g2) 3 3 3 ⇒ (g1g2) = g1g2 3 3 ⇒ g1g2g1g2g1g2 = g1g2 2 2 2 2 ⇒ g2g1g2g1 = g1g2 = g2g1 ( using(i)) ⇒ g g = g g 2 1 1 2 ∴ G is abelian so every subgroup of G is normal in G. Hence G Dedekind group. 4.Conclusion In this paper I have discussed mostly those cases for the Dedekind group which are directly linked with the commutative group but there are also Dedekind group which are non commutative known as Hamiltonian group. The scope of this paper to generalise these results or properties of Dedekind group for more deductions which are useful for the development of the theory of Dedekind groups. References [1] Dewayne S. Nymann, Dedekind Groups, Pacific Journal of Mathematics Vol.
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