Define a group. Give a suitable example.
A non-empty set G together with a binary operation * is called a group if it satisfies:
- Closure: a * b is in G for all a, b in G.
- Associativity: (a * b) * c = a * (b * c) for all a, b, c in G.
- Identity: There exists e in G such that a * e = e * a = a.
- Inverse: For every a in G, there exists a' in G such that a * a' = a' * a = e.
Example: The set of integers Z under the operation of addition (+) is a group.
Write the Cayley table for the set G = {0, 1, 2, 3} with respect to usual multiplication. Is it a group?
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 |
| 2 | 0 | 2 | 4 | 6 |
| 3 | 0 | 3 | 6 | 9 |
Conclusion: No, it is not a group.
Specifically, the element 0 does not have a multiplicative inverse, and the set is not closed under usual multiplication (e.g., 2 * 2 = 4, which is not in G).Define order of an element of a group. Find the order of 2 in Z₄.
The order of an element 'a' in a group G is the smallest positive integer n such that aⁿ = e (where e is the identity). In additive notation, this is na = 0.
In Z₄ = {0, 1, 2, 3} under addition modulo 4:
Final Answer: The order of 2 in Z₄ is 2.
(a) Show that the set GL(2, R) of 2x2 matrices with real entries and non-zero determinant is a group w.r.t. matrix multiplication.
To prove GL(2, R) is a group:
(b) Show that a non-empty subset H of a group G is a subgroup of G if and only if ab⁻¹ is in H whenever a, b are in H.
Proof:
(c) Prove the uniqueness of identity and uniqueness of inverses in a group.
Uniqueness of Identity: Suppose there are two identities e and e'. Then e * e' = e (if e' is identity) and e * e' = e' (if e is identity). Thus e = e'.
Uniqueness of Inverses: Let 'a' have two inverses b and c. Then ab = e and ac = e. Now, b = be = b(ac) = (ba)c = ec = c. Thus b = c.
(d) Let G = {[[a, a], [a, a]] | a ∈ R, a ≠ 0}. [span_19](start_span)Show that G is a group w.r.t. multiplication.[span_19](end_span)
Let A = [[a, a], [a, a]] and B = [[b, b], [b, b]].
Define centralizer of an element of a group.
The centralizer of an element 'a' in a group G, denoted C(a), is the set of all elements in G that commute with 'a':
C(a) = {x ∈ G | xa = ax}
Let G = <a> be a cyclic group of order 25. Justify if a⁵ is a generator of G.
For a cyclic group G = <a> of order n, aᵏ is a generator if and only if gcd(k, n) = 1.
Here, n = 25 and k = 5. Since gcd(5, 25) = 5 ≠ 1, a⁵ is not a generator of G.
(a) Show that the center of a group is a subgroup of the group.
The center Z(G) = {z ∈ G | zg = gz for all g ∈ G}.
(b) Let G = <a> be a cyclic group of order n. Prove that G = <aᵏ> iff gcd(k, n) = 1.
Proof: The order of element aᵏ is |aᵏ| = n / gcd(k, n). For aᵏ to be a generator, its order must equal the order of the group, which is n. Thus, n / gcd(k, n) = n, which implies gcd(k, n) = 1.
(c) Justify if the center of a group is always Abelian. Justify if the centralizer of an element is always Abelian.
(d) Show that every subgroup of a cyclic group is cyclic.
Let G = <a> and H be a subgroup. If H = {e}, it is cyclic. If H ≠ {e}, let m be the smallest positive integer such that aᵐ ∈ H. We claim H = <aᵐ>. For any aⁿ ∈ H, by division algorithm n = mq + r where 0 ≤ r < m. Then aʳ = aⁿ(aᵐ)⁻ᵝ. Since aⁿ, aᵐ ∈ H, aʳ must be in H. Since m is the smallest positive integer, r must be 0. Thus n = mq, and aⁿ = (aᵐ)ᵝ.
Define symmetric group Sₙ.
The symmetric group Sₙ is the group of all permutations (one-to-one and onto functions) of a finite set of n elements under the operation of function composition.
(a) Show that every permutation of a finite set can be written as a product of disjoint cycles. Express (12)(45)(153)(24) as disjoint cycles.
Decomposition: Starting with 1 → 5 → 4 → 2 → 1. Next available 3 → 3. Next available 5 (already used). Result: (1542)(3) or simply (1542).
(b) Show that disjoint cycles commute.
Let α and β be disjoint cycles. They move disjoint sets of elements. If an element x is moved by α, it is fixed by β. Thus αβ(x) = α(x) and βα(x) = β(α(x)) = α(x) (since α(x) is also fixed by β). If x is moved by neither, both sides fix it. Thus αβ = βα.
State and prove Lagrange's theorem.
Theorem: The order of every subgroup H of a finite group G divides the order of G.
Proof: Let |G| = n and |H| = m. The distinct left cosets of H partition G. Since each coset has exactly m elements and there are, say, k distinct cosets, then n = mk. This implies m divides n, i.e., |H| divides |G|.
Show that a subgroup H of G is normal iff xHx⁻¹ ⊆ H for all x ∈ G.
Proof: A subgroup is normal if xH = Hx for all x. This is equivalent to xHx⁻¹ = H. The condition xHx⁻¹ ⊆ H for all x implies that for x⁻¹, x⁻¹H(x⁻¹)⁻¹ ⊆ H, which means x⁻¹Hx ⊆ H, or H ⊆ xHx⁻¹. Thus xHx⁻¹ = H.
Give example, with justification, of an integral domain that is not a field.
Example: The ring of integers Z.
Justification: Z has no zero divisors (if ab = 0, then a=0 or b=0), so it is an integral domain. However, elements like 2 have no multiplicative inverse in Z (1/2 is not an integer), so it is not a field.
Prove that every non-zero finite integral domain is a field.
Proof: Let D be a finite integral domain with elements {x₁, x₂, ..., xₙ}. For any non-zero 'a' in D, consider the set {ax₁, ax₂, ..., axₙ}. Since D has no zero divisors, all these products are distinct. Thus, one of them must be the unity '1'. So axᵢ = 1 for some xᵢ, meaning 'a' has an inverse. Hence, D is a field.