MTH405 MidTerm Quiz by taleemi markaz 2024
Автор: TALEEMI MARKAZ
Загружено: 2024-02-05
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Описание:
MTH405 MidTerm Quiz by taleemi markaz 2024 Part 1
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Questions:
𝒂∗𝑯=𝑯∗𝒂 relation holds if _______
The operation of addition '+' satisfies the commutative property
The set of nth roots of unity is a cyclic group under
If a,b are the elements of group G, then (𝒃^(−𝟏) 𝒂^(−𝟏) )^(−𝟏)=____
The set {1,-1} is a group under addition.
If X has 4 element, then total number of permutations on X is _______
Subtraction is a binary operation on the set of Real numbers R.
The group (Z,+) is a finite group
If G is an abelian group and H is subset of G, then 𝒂𝑯=𝑯𝒂 for some a in G.
If X consists of the elements 1,2,....7, then the symbol (1,3,4,2,6) means the permutation ______
If 𝒇=(■8(𝟏&𝟐&𝟑@𝟐&𝟏&𝟑)) 𝒂𝒏𝒅 𝒈=(■8(𝟏&𝟐&𝟑@𝟐&𝟑&𝟏)) are two permutations, then their product permutation, 𝒇𝒈=(■8(𝟏&𝟐&𝟑@𝟑&𝟏&𝟐))
Under multiplication, 1 is not an idempotent element
A non-empty set with the ____ binary operation is Semigroup.
A non-empty set with a binary operation is called _____
The set of real numbers is an abelian group under
A cycle of length ____ is called the transposition.
Every cyclic permutation can be expressed as a _____ of transpositions.
The length of cyclic permutation (■8(𝒂_𝟏&𝒂_𝟐&𝒂_𝟑@𝒂_𝟐&𝒂_𝟑&𝒂_𝟏 )) is
The set G={1,-1,i,-i} is a NOT group under multiplication.
The group {1,-1} is subgroup of the group {1,-1,i,-i}
The set of natural numbers N starts from 1.
A group G is abelian if and only if (𝒂𝒃)^𝟐=𝒂^𝟐 𝒃^𝟐 for all 𝒂, 𝒃∈𝑮
The length of the cyclic permutation is
(■8(𝟐&𝟑&𝟏&𝟒&𝟓&𝟔@𝟏&𝟑&𝟓&𝟒&𝟐&𝟔))
If X has 4 element, then total number of permutations on X is _______
If index of H in G is 4, then there are 4 _________
If index of H in G is 4, then there are 4 _________
The operation of addition '+' satisfies the commutative property
Is the following true or false?
(■8(𝟏&𝟐&𝟑@𝟐&𝟑&𝟏))(■8(𝟏&𝟐&𝟑@𝟑&𝟏&𝟐))=(■8(𝟏&𝟐&𝟑@𝟏&𝟐&𝟑))
The group (G,*) is said to be an abelian group or commutative group if for all 𝒂,𝒃∈𝑮, 𝒂∗𝒃=
Let order of a group be 35. Which of the following CANNOT be the order of the subgroup of G?
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