Problem on Finite Abelian Groups: Part-II
Welcome to our comprehensive exploration of Finite Abelian Groups! In this article, we present a curated collection of intriguing problems that not only test your understanding of group theory fundamentals but also help you master essential techniques. Whether you are a student gearing up for examinations or a mathematics enthusiast passionate about abstract algebra, our detailed examples—ranging from class equations to analyses of group centers—will enrich your mathematical insight. Embrace the challenge and discover the elegance behind the structure of finite groups.
Problem 1
Does there exist a group whose class equation is \[ \pmb{10 = 1 + 1 + 1 + 2 + 5?} \]
Answer: Assume \( G \) is a group of order 10. The class equation indicates that there are three singleton conjugacy classes, suggesting that the center \( Z(G) \) comprises three elements (i.e. \(|Z(G)| = 3\)). However, by Lagrange’s Theorem, the order of any subgroup must divide the group order, and since 3 does not divide 10, such a group cannot exist.
Problem 2
Is it possible for a group to have the class equation \[ \pmb{14 = 1 + 1 + 1 + 2 + 2 + 7 ?}\]
Answer: Let \( G \) be a group of order 14. This equation reveals three conjugacy classes of size 1, indicating that \( Z(G) \) must contain exactly three elements. Since any subgroup’s order (including that of the center) must divide 14, and 3 is not a divisor of 14, no group with this class equation can exist.
Problem 3
Can there be a group with the class equation \[ \pmb{15 = 1 + 1 + 3 + 5 + 5 ?}\]
Answer: Suppose \( G \) is a group of order 15. The two singleton terms in the equation imply that \( Z(G) \) contains 2 elements. However, by Lagrange’s Theorem the order of the center must divide 15, and since 2 is not a divisor of 15 (its divisors being 1, 3, 5, and 15), no such group exists.
Problem 4
Does a group exist with the class equation \[ \pmb{18 = 1 + 1 + 1 + 1 + 2 + 3 + 9? }\]
Answer: Let \( G \) be a group of order 18. The four singleton conjugacy classes indicate that the center \( Z(G) \) would have 4 elements. However, the divisors of 18 are 1, 2, 3, 6, 9, and 18, and since 4 does not appear among these divisors, such a group is impossible.
Problem 5
Can a group exist with the class equation \[ \pmb{20 = 1 + 1 + 1 + 2 + 5 + 10 ?}\]
Answer: Assume \( G \) is a group of order 20. The three singleton elements indicate that \( Z(G) \) has 3 elements. However, since the divisors of 20 are 1, 2, 4, 5, 10, and 20, and 3 is not among them, no group of order 20 can have this class equation.
Problem 6
Is there a group whose class equation is \[ \pmb{21 = 1 + 1 + 1 + 1 + 3 + 7 + 7?} \]
Answer: Let \( G \) be a group of order 21. The four singleton terms imply that \( Z(G) \) consists of 4 elements. Since the order of any subgroup must divide 21, and the divisors of 21 are 1, 3, 7, and 21, the number 4 is not permissible. Consequently, such a group cannot exist.
Problem 7
Does there exist a group with the class equation \[ \pmb{22 = 1 + 1 + 1 + 1 + 1 + 11 + 2 + 2 + 2? }\]
Answer: Suppose \( G \) is a group of order 22. The five instances of singleton conjugacy classes indicate that \( Z(G) \) has 5 elements. However, because the order of \( G \) is 22 and its divisors are only 1, 2, 11, and 22, the number 5 cannot be accommodated as a subgroup order. Therefore, no such group exists.
Problem 8
Can a group have the class equation \[ \pmb{26 = 1 + 1 + 1 + 13 + 2 + 2 + 2 + 2 + 2 ?}\]
Answer: Let \( G \) be a group of order 26. Here, the three singleton classes imply that \( Z(G) \) must contain 3 elements. As 3 does not divide 26 (whose divisors are 1, 2, 13, and 26), such a configuration is impossible.
Problem 9
Does there exist a group whose class equation is \[ \pmb{28 = 1 + 1 + 1 + 7 + 7 + 7 + 4 ?}\]
Answer: Assume \( G \) is a group of order 28. The three occurrences of a singleton conjugacy class suggest that \( Z(G) \) consists of 3 elements. However, since the divisors of 28 are 1, 2, 4, 7, 14, and 28, and 3 is not among these, no such group exists.
Problem 10
Can a group exist with the class equation \[ \pmb{30 = 1 + 1 + 1 + 1 + 6 + 10 + 10 ?}\]
Answer: Let \( G \) be a group of order 30. The four singleton entries in the class equation imply that the center \( Z(G) \) must have 4 elements. Given that the divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30, and since 4 is not a divisor, it follows that such a group cannot exist.
Conclusion
In summary, this collection of problems underscores the fundamental relationship between group order, centers, and class equations in group theory. By analyzing these examples, you gain valuable insights into the structural intricacies of finite groups, particularly Finite Abelian Groups. Continue your journey with us to master these concepts and refine your problem-solving skills in group theory.
Related Docs
FAQs
Inner Product Space
- What is an inner product space?
An inner product space is a vector space equipped with an additional structure called an inner product. The inner product allows for the definition of geometric concepts such as length, angle, and orthogonality.
- What is an inner product?
An inner product is a function that takes two vectors from the vector space and returns a scalar, typically denoted as ( langle u, v rangle ) for vectors ( u ) and ( v ). This function must satisfy certain properties: linearity in the first argument, symmetry, and positive-definiteness.
- What are the properties of an inner product?
- Linearity in the first argument:** ( langle au + bv, w rangle = a langle u, w rangle + b langle v, w rangle ) for all scalars ( a, b ) and vectors ( u, v, w ).
- Symmetry:** ( langle u, v rangle = langle v, u rangle ) for all vectors ( u, v ).
- Positive-definiteness:** ( langle u, u rangle geq 0 ) for all vectors ( u ), and ( langle u, u rangle = 0 ) if and only if ( u ) is the zero vector.
- How does the inner product relate to the norm of a vector?
The norm (or length) of a vector ( u ) in an inner product space is defined as the square root of the inner product of the vector with itself, i.e., ( |u| = sqrt{langle u, u rangle} ).
- What is orthogonality in an inner product space?
Two vectors ( u ) and ( v ) are orthogonal if their inner product is zero, i.e., ( langle u, v rangle = 0 ). Orthogonality generalizes the concept of perpendicularity in Euclidean space.
- What is the Cauchy-Schwarz inequality?
The Cauchy-Schwarz inequality states that for all vectors ( u ) and ( v ) in an inner product space, ( |langle u, v rangle| leq |u| |v| ). This inequality is fundamental in the study of inner product spaces.
- What is an orthonormal basis?
An orthonormal basis of an inner product space is a basis consisting of vectors that are all orthogonal to each other and each have unit norm. This means that for an orthonormal basis ( {e_1, e_2, ldots, e_n} ), ( langle e_i, e_j rangle = 1 ) if ( i = j ) and ( 0 ) otherwise.
- How do you project a vector onto another vector in an inner product space?
The projection of a vector ( u ) onto a vector ( v ) is given by ( left(frac{langle u, v rangle}{langle v, v rangle}right) v ). This formula uses the inner product to find the scalar component of ( u ) in the direction of ( v ).
- What is the Gram-Schmidt process?
The Gram-Schmidt process is a method for orthonormalizing a set of vectors in an inner product space. Given a set of linearly independent vectors, the process constructs an orthonormal set of vectors that spans the same subspace.
