Abstract Algebra: An Introduction (3rd Edition)
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The identity element will belong to since is a ... more
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; is the dihedral group and is the subgroup ... more
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; is the dihedral group and is the subgroup ... more
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; The right coset of in can be defined as . ... more
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; is the multiplicative subgroup of and is the ... more
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; Since the order of is 16 where is the ... more
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; The order of group is 8 where is the dihedral ... more
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; The order of group is 12 where is the group of ... more
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1 ; The order of group is 20 where is the group of... more
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; The order of group is 40 where is the group of ... more
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; The order of is 24 which is where is the ... more
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; Order of is 15, and that of is 5. ; is ... more
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27720 ; Order of every element must divide the ... more
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50 ; Since, group G has subgroups of order 10 and ... more
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and are subgroups of , so their intersection is ... more
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and Let Since and are ... more
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Let Since, G does not have non trivial proper ... more
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Since the group has order 25 and order of every ... more
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2 ; Cardinality of a is 30. Finding the ... more
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Since, the order of group is 8 and every element ... more
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For , as . , Here so implies that in U(n). ; ... more
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Let As , so Solving , Since , implies that in .As... more
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G is finite and , andAlso, ; Using theorem that if... more
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Every element has its inverse which has the same ... more
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Abelian group contains one element of order 2, be ... more
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By Lagrange theorem, the group does not contain ... more
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If say be a proper subgroup of a group then the ... more
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First, use Lagrange Theorem to establish a ... more
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For any the function is a group homomorphism as ... more
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Assume group to not contain the identity element ... more
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There must exist at least one element say of order... more
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First assume the case where the group does not ... more
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Assume to be a subgroup generated by then . Now, ... more
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Assume to be a subgroup generated by then . Now, ... more
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Identity element belongs to as is a subgroup such ... more
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Since the order of is 8 where is the dihedral ... more
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Left cosets of obtained using the multiplication ... more
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Let , so the elements of are only.Let and , ... more
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The permutation group and,Simplifying the ... more
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Let be two elements.Now, let and be an arbitary... more
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Since, is a subgroup of and for all such that .... more
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Let and So for every ,As elements of center ... more
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Let be the characteristic subgroup of .Let , for ... more
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Let be an automorphism and let Taking, So, and ... more
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Proving the first part Let be a group and... more
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Let The permutation group andHere, index of in... more
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Let , is even as property of , Using these ... more
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From the given data, be the normal subgroup of ... more
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Here, , so Let Since, is homomorphismNow take, So... more
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Let are two normal subgroups of .Let Using the ... more
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Intersection of two subgroups are also a subgroup.... more
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for all and . ; To prove that for all and it... more
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is normal subgroup of ; Since, is a surjective... more
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is a normal subgroup of . ; To prove that is a ... more
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is a normal subgroup of . ; Let and .To prove ... more
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is a normal subgroup of . ; Let be the unique ... more
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be the normal subgroup of if and only if it has ... more
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Cyclic subgroup is the normal subgroup of if ... more
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is a normal subgroup of . ; To prove that is a ... more
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Define the map defined by . Now to show is a ... more
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To show is a normal subgroup of , it is ... more
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To show is a subgroup of , it is sufficient to ... more
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Since, Now, the only thing left to prove is for ... more
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To show is a normal subgroup of , it is ... more
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For . The mapping with is a automorphism of as... more
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For , consider the subgroupIt follows from the ... more
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4 ; Let be the subgroup of generated by 4. Then ... more
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5 ; Let be the order of in the group such that ... more
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Now, since the identity element in is so is the ... more
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is a group which consists of two elements as is a ... more
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is a cyclic subgroup generated by 6 thus the order... more
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is a subgroup which consists of two elements so ... more
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is a cyclic subgroup generated by 5 thus the order... more
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is a cyclic subgroup generated by thus the order ... more
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is a cyclic subgroup generated by thus the order ... more
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Every subgroup of the group is a normal subgroup ... more
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Assume a nonidentity element say which belongs to ... more
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; Consider an element of order that belongs to ... more
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; Consider an element of order that belongs to ... more
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Define the mapping as follows: To show is well ... more
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Let be an arbitrary element in . Since has the ... more
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Since and only group of order 4 are either or , ... more
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If possible, suppose that there is a non-identity ... more
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Define the mapping by .Now to show is isomorphic... more
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; To describe the group , claim is that the group... more
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Since is cyclic. So, there exists an element ... more
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Set of elements of finite order in the group is ... more
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is isomorphic to a subgroup of . ; To prove ... more
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Every element of has finite order. ; It is given ... more
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also has an element of order . ; It is given that... more
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Centre of is either or . ; It is given that the... more
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is finitely generated. ; It is given that be a ... more
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The element , since . Consider the elements and ... more
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As the normal subgroup of is and as proved ... more
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To prove that is isomorphic to : The mapping from... more
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Proof:The mapping from such that where ... more
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Assume . To show given mapping it is a ... more
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Assume . To show given mapping is a homomorphism... more
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Assume . To show given mapping is a homomorphism... more
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Assume . To show given mapping is a homomorphism... more
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Assume To show given mapping it is a homomorphism... more
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Assume . To show given mapping is a homomorphism... more
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Assume . To show given mapping is a homomorphism... more
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Prove the given statement by assuming , ... more
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The function is defined as . Now it proved that ... more
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Firstly we will show h is well defned i.e. we ... more
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Firstly we will show h is well defined i.e. we ... more
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Let G and H be a group it is given that is an ... more
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Firstly we will show that f is homomorphism we ... more
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Homomorphic images of are quotients group of by ... more
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Homomorphic images of are quotients of by normal... more
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(First Isomorphism Theorem). Let be onto ... more
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Let us suppose so which means and we need to ... more
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let us suppose we will show f is homomorphismhence... more
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Let us define a map is onto map since for every ... more
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Let us define f is onto as for every element in ... more
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(First Isomorphism Theorem). Let be onto ... more
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(First Isomorphism Theorem). Let be onto ... more
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Define the mapping by . Firstly show that is a ... more
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As is a subgroup of , and if a group has finite ... more
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Define the mapping by Now to prove is a ... more
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To show the mapping is well defined. Let be such... more
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Since is a surjective homomorphism from to with... more
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Since and is given in the question.Define a ... more
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Let be the normal subgroup of .Define the ... more
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The group is the group of all rational numbers.... more
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Necessary condition,Let be a simple group.To ... more
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Define a mapping by Now the aim is to show that ... more
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; Since the center of the group is defined as ... more
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for ; Suppose contains a permutation which is ... more
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Suppose where and for all .Since every ... more
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Proof this statement by contradiction.Suppose that... more
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Since the order of a permutation which is ... more
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Suppose be a subgroup of order 2 in for . Since... more
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If then proof is done.So, let . Now the aim is to... more
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If then proof is done.So let . Then the aim is to... more
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Let be a subgroup of whose index is 2 in . Now ... more
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By the property of homomorphism, the image of the ... more