Practical Questions1 Prove That Every Subgroup H Of Cyclic Group G I ✓ Solved

PRACTICAL QUESTIONS (1) Prove that every subgroup H of cyclic group G is a cyclic group. Find all generators of group Z32. (2) Show that group Un (n th unit root) and group Zn are isomorphic. (3) Is all groups of order 6 commutative ? If not commutative, give an example. How about groups of order 5? (Are they always commutative). (4) Let set G be the Cartisian product of G1 à—G2, where G1 and G2 are 2 groups. Define a binary operation ∗ on G by (a, b) ∗ (c, d) = (ac, bd).

Show that (G,∗) is group. If both G1 and G2 are commutative groups, show G is also a commutative group. (5) Let H := {A ∈ Mnà—n|A = AT , det(A) 6= 0} be a set, ∗ be matrix multiplication. Is the set (H,∗) a group? What if ∗ is the matrix addition? (6) Assume that G is a group such that for all x ∈ G, x∗x = e. Prove that G is an abelian group.

Date: April 23, 2021. 1

Paper for above instructions

Introduction
Group theory is a branch of mathematics that studies the algebraic structures known as groups. A group is defined as a set combined with an operation that satisfies the four fundamental properties: closure, associativity, identity, and invertibility. This paper examines six practical questions based on group theory, specifically focusing on cyclic groups, isomorphisms, commutativity, the Cartesian product, matrix groups, and properties that lead to abelian groups.

Question 1: Subgroups of Cyclic Groups

Statement

To prove that every subgroup \( H \) of a cyclic group \( G \) is cyclic.

Proof

Let \( G \) be a cyclic group generated by an element \( g \) (i.e., \( G = \langle g \rangle \)). This means every element in \( G \) can be expressed in the form \( g^n \) for some integer \( n \).
Let \( H \) be a subgroup of \( G \). Since \( H \) is a subgroup, it must also be non-empty and must contain the identity element \( e \) (which is \( g^0 = e \)). Since \( g \) generates \( G \), there exists an integer \( k \) such that \( h \in H \) can be expressed as \( h = g^m \) for some integer \( m \).
Now consider the set of all such integers \( n \) for which \( g^n \in H \). Because \( H \) is closed under the operation, the subgroup generated by \( g^m \) is given by:
\[
\langle g^m \rangle = \{ g^{mq} : q \in \mathbb{Z} \}
\]
Thus, \( H \) is generated by some element, specifically \( g^{d} \), where \( d \) is the greatest common divisor of the orders of the elements in \( H \).
Hence, any subgroup \( H \) of cyclic group \( G \) is cyclic, proving the statement is true.

Generators of \( \mathbb{Z}_{32} \)

The group \( \mathbb{Z}_{32} \) can be generated by any element that is coprime to 32. The totient function \( \phi(32) = 32(1 - 1/2)(1 - 1/4) = 16 \) indicates there are 16 generators. The integers that are coprime to 32 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, and 31.
Each of these elements can serve as a generator for the group \( \mathbb{Z}_{32} \).

Question 2: Isomorphism Between \( U(n) \) and \( \mathbb{Z}_n \)

Statement

Show that group \( U(n) \) (the group of units mod \( n \)) and group \( Z_n \) are isomorphic.

Proof

The group \( U(n) \) consists of all integers \( k \) such that \( 1 \leq k < n \) and \( \text{gcd}(k, n) = 1 \). The structure of \( U(n) \) can have an additive structure similar to \( Z_n \) through the operation of multiplication modulo \( n \).
To show they are isomorphic, we define a homomorphism:
\[
f: \mathbb{Z}_n \rightarrow U(n)
\]
where \( f(k) = a^k \mod n \), for a generator \( a \) of \( U(n) \). Since every element \( k \in \mathbb{Z}_n \) corresponds to \( a^k \), the mapping preserves the group operation, thus demonstrating that there exists an isomorphism between \( U(n) \) and \( \mathbb{Z}_n \).

Question 3: Commutativity of Groups of Order 6 and 5

Statement

Determine if all groups of order 6 are commutative.

Answer

No, not all groups of order 6 are commutative. The symmetric group \( S_3 \) (the group of permutations of three elements) is a non-abelian group of order 6.

Groups of Order 5

On the other hand, any group of prime order \( p \) (in this case 5) is always cyclic (and therefore abelian). This is due to the fact that any non-identity element must generate the group itself.

Question 4: Cartesian Products of Groups

Statement

Let \( G \) be the Cartesian product \( G_1 \times G_2 \), with operations defined component-wise as:
\[
(a, b) * (c, d) = (ac, bd)
\]

Proof

Closure: For \( (a, b), (c, d) \in G\), both \( ac \) and \( bd \) belong to their respective groups.
Associativity: From the definition of multiplication within groups, \((ab)c = a(bc)\).
Identity: Let \( e_1 \) be the identity of \( G_1 \) and \( e_2 \) that of \( G_2\). Then \( (e_1, e_2) \) acts as the identity in \( G \).
Inverses: Each element \( (a, b) \) has an inverse \( (a^{-1}, b^{-1}) \).

Commutativity of \( G \)

If both \( G_1 \) and \( G_2 \) are abelian, then for any two elements in \( G \), we have:
\[
(a, b)(c, d) = (ac, bd) = (ca, db) = (c, d)(a, b)
\]

Conclusion

Thus \( G \) remains an abelian group.

Question 5: Matrix Groups

Statement

Let \( H = \{ A \in M_n | A = A^T, \text{det}(A) \neq 0 \} \).

Proof

Matrix Multiplication:
- Identity: The identity matrix \( I \) is symmetric and has a non-zero determinant.
- Closure: The product of two symmetric matrices is symmetric; thus, if \( A, B \in H \), then \( AB \in H \).
- Inverses: The inverse of a non-singular matrix is also non-singular.
Thus, \( (H, \cdot) \) forms a group.
Matrix Addition:
The sum of two symmetric matrices is symmetric, but the sum does not necessarily yield a determinant non-zero outcome, hence does not form a group under addition.

Question 6: Groups Where \( x * x = e \)

Proof

Let \( G \) be a group where for all \( x \in G \), we have \( x * x = e \).
For any elements \( a, b \in G \):
- \( a * b = e \) implies \( b = a^{-1} \),
- \( a b = b a \) since \( abb = eb = ae \).
Thus \( G \) is abelian.

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