# MTH501 Assignment Solution 2023

MTH501 Assignment Solution 2023. Mathematics is the study of patterns and structures. In MTH501, we focus on abstract algebra, which studies algebraic structures such as groups, rings, and fields. This assignment will test your understanding of the concepts and techniques covered in the course.

Question 1: Prove that every finite integral domain is a field.

Solution: Let R be a finite integral domain. Since R is finite, every non-zero element of R is invertible. Thus, R is a division ring. To show that R is a field, we need to show that every non-zero element has a multiplicative inverse.

Suppose a is a non-zero element of R. Consider the sequence a, a^2, a^3, …, which must eventually repeat since R is finite. Let n be the smallest positive integer such that a^n = a^m for some m < n. Then a^n – a^m = 0, so a^m(a^(n-m) – 1) = 0. Since R is an integral domain, either a^m = 0 or a^(n-m) – 1 = 0. But a^m cannot be zero since a is non-zero, so we must have a^(n-m) = 1. Thus, a^(n-m-1) is the inverse of a, and so a is invertible. Therefore, R is a field.

Question 2: Let G be a finite group and let H be a subgroup of G. Show that |H| divides |G|.

Solution: Let G be a finite group and let H be a subgroup of G. Consider the left cosets of H in G, denoted by gH for g in G. Since each element of G belongs to precisely one left coset and the left cosets partition G, we have |G| = [G: H] |H|, where [G: H] denotes the number of left cosets of H in G.

To show that |H| divides |G|, it suffices to show that [G: H] is an integer. We claim that the left cosets of H in G form a partition of G, which implies that [G: H] is the number of cosets.

To prove this claim, suppose gH and g’H are distinct left cosets of H in G. Then there exists h in H such that gh is not equal to g’h. But then (g’h)^-1 gh is an element of H, which implies that g’^-1 g is an element of H. Therefore, gH and g’H cannot be disjoint, and so they must either be equal or overlap.

Since the left cosets of H in G form a partition of G, we have [G:H] = |G|/|H|, which implies that |H| divides |G|.

### MTH501 Assignment Solution 2023

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