For students pursuing advanced mathematics, tackling intricate questions is crucial for deepening understanding and mastering the subject. In this blog, we delve into two challenging problems commonly encountered at the master's level and provide detailed solutions. If you find yourself needing further assistance with similar problems, mathsassignmenthelp.com offers expert help with assignments. Whether it's for discrete math or other areas, you can request Do My Discrete Math Assignment services to get tailored support.

Proving Every Subgroup of an Abelian Group is Normal

Question: In an abelian group G, show that every subgroup of G is normal.

Answer: To demonstrate that every subgroup of an abelian group G is normal, we first need to understand the concept of a normal subgroup. A subgroup H of G is considered normal if for every element g in G and every element h in H, the result of the operation g multiplied by h multiplied by the inverse of g is still within H.

Since G is abelian, it means that the group operation is commutative. In other words, for any elements g and h in G, the order in which you perform the operation does not affect the result, so g multiplied by h is the same as h multiplied by g.

Now, consider any subgroup H of G. To show that H is normal, we need to prove that when we conjugate an element h in H by any element g in G, the result is still in H. Conjugating h by g involves multiplying h on the left and right by g.

In an abelian group, because the operation is commutative, the conjugation of h by g simply returns h itself. This is because g multiplied by h is the same as h multiplied by g, and the operation g multiplied by the inverse of g cancels out, leaving just h.

Since the conjugated element is still h, which is an element of H, it confirms that H is a normal subgroup of G. Therefore, every subgroup of an abelian group is normal by this property.

Demonstrating the Fundamental Theorem of Algebra

Question: Prove that every polynomial with complex coefficients, which is not a constant, has at least one complex root.

Answer: The Fundamental Theorem of Algebra states that every polynomial with complex coefficients, provided it is not constant, has at least one complex root. To prove this, we use the concept of continuity and the completeness of the complex number system.

Consider a polynomial function of degree n with complex coefficients. If this polynomial had no roots, it would mean that the polynomial never takes the value zero in the complex plane. However, polynomial functions are continuous, and in the field of complex numbers, they are also bounded. This means that if a polynomial did not have any roots, it would imply that the polynomial function must be constant, which contradicts the assumption that it is not constant.

The continuity of polynomial functions in the complex plane ensures that these functions will cross the horizontal axis (i.e., take the value zero) at least once if they are not constant. The fact that complex numbers form a complete field guarantees that such a root exists.

Thus, by understanding the nature of polynomial functions and their behavior in the complex plane, we conclude that every non-constant polynomial must have at least one complex root. This root can be found through various algebraic methods or factoring techniques.

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