For students delving into higher-level mathematics, tackling complex questions can be both challenging and rewarding. In this blog, we address two intricate mathematical problems and provide comprehensive solutions. For additional support with your mathematical studies, including assistance with accounting-related questions, visit mathsassignmenthelp.com, where you can find expert Discrete Math Assignment Help Online for tailored to your needs.

Understanding Subgroup Normality in Abelian Groups

Question: In an abelian group, prove that every subgroup is normal.

Answer: To demonstrate that every subgroup within an abelian group is normal, we first need to grasp the concept of a normal subgroup. A subgroup is deemed normal if, for every element in the larger group, conjugating any element of the subgroup by this larger group’s element results in an element that remains within the subgroup.

In an abelian group, the key feature is that the group operation is commutative. This means that the order in which you perform the group operations does not affect the result. Specifically, for any elements $g$ and $h$ in the group, the result of combining $g$ and $h$ in any order will be the same.

Given this commutativity, let’s consider any subgroup $H$ of an abelian group $G$. To verify that $H$ is normal, we need to check that for every element $g$ in $G$ and every element $h$ in $H$, the element ghg−1g h g^{-1}ghg−1 is still in $H$.

In an abelian group, because the operation is commutative, conjugating $h$ by $g$ simplifies to just $h$. This is because: ghg−1=hg h g^{-1} = hghg−1=h

Thus, since the conjugated element ghg−1g h g^{-1}ghg−1 is simply $h$, and $h$ is in $H$, it follows that $H$ is invariant under conjugation. Hence, every subgroup of an abelian group is indeed normal.

Proving the Fundamental Theorem of Algebra

Question: Show that every polynomial with complex coefficients and degree greater than zero has at least one complex root.

Answer: The Fundamental Theorem of Algebra states that every polynomial with complex coefficients has at least one root within the complex number system. This theorem is crucial in understanding the behavior of polynomials in complex analysis.

To prove this, we utilize the completeness of the complex numbers. Consider a polynomial function of degree $n$ with complex coefficients. If this polynomial were to have no roots, it would imply that the polynomial function never crosses the horizontal axis in the complex plane.

However, by the nature of complex polynomials, which are continuous functions, the polynomial's values cannot be unbounded or avoid zero. If a polynomial function did not have any roots, it would mean that the function is either always positive or always negative, which is impossible for non-constant polynomials due to their continuous nature.

Thus, by the completeness of the complex numbers and their properties, every polynomial of degree greater than zero must intersect the horizontal axis at least once. This intersection represents a root of the polynomial in the complex number system.

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