Representative of subgroup

Question

If H is a subgroup of G, and the index of H is finite (G:H = n < ∞). Prove that there are elements a1, a2, ... an that can be left and right representatives of H cosets. That is G = UaiH and G=UHai

Answer 1

Proof

1. Existence of Left Coset Representatives: Since the index of H in G is finite (G:H = n), there are exactly 'n' distinct left cosets of H in G. Let these distinct left cosets be: a₁H, a₂H, ..., a<0xE2><0x82><0x99>H where a₁, a₂, ..., a<0xE2><0x82><0x99> are elements of G. These elements aᵢ are the left coset representatives.
2. By the definition of cosets, the union of all distinct left cosets partitions the group G. That is: G = a₁H ∪ a₂H ∪ ... ∪ a<0xE2><0x82><0x99>H This can be written more concisely using the union notation: G = ⋃ᵢ<0xE2><0x82><0x99>₁ aᵢH
3. Existence of Right Coset Representatives: Similarly, since the index of H in G is 'n', there are exactly 'n' distinct right cosets of H in G. Let these distinct right cosets be: Hb₁, Hb₂, ..., Hb<0xE2><0x82><0x99> where b₁, b₂, ..., b<0xE2><0x82><0x99> are elements of G. These elements bᵢ are the right coset representatives.
4. The union of all distinct right cosets also partitions the group G: G = Hb₁ ∪ Hb₂ ∪ ... ∪ Hb<0xE2><0x82><0x99> = ⋃ᵢ<0xE2><0x82><0x99>₁ Hbᵢ
5. Connecting Left and Right Representatives: We need to show that there exists a set of elements that can serve as both left and right representatives.
6. Consider the set of left coset representatives we found: {a₁, a₂, ..., a<0xE2><0x82><0x99>}. We will show that the set of inverses of these elements, {a₁⁻¹, a₂⁻¹, ..., a<0xE2><0x82><0x99>⁻¹}, can serve as a set of right coset representatives.
7. Let's look at a right coset formed by one of these inverses, say Haᵢ⁻¹. We want to show that this right coset is one of the distinct right cosets of H.
8. We know that the mapping φ: aH → Ha⁻¹ is a bijection (one-to-one and onto) between the set of left cosets and the set of right cosets. This is a standard result in group theory.
9. Since the left cosets a₁H, a₂H, ..., a<0xE2><0x82><0x99>H are all distinct, their corresponding right cosets under the bijection, Ha₁⁻¹, Ha₂⁻¹, ..., Ha<0xE2><0x82><0x99>⁻¹, must also be distinct.
10. Furthermore, since there are exactly 'n' distinct right cosets, the set {Ha₁⁻¹, Ha₂⁻¹, ..., Ha<0xE2><0x82><0x99>⁻¹} must be the complete set of distinct right cosets.
11. Therefore, the elements a₁⁻¹, a₂⁻¹, ..., a<0xE2><0x82><0x99>⁻¹ can serve as right coset representatives.
12. Conclusion: We have shown that if H is a subgroup of G with a finite index 'n', we can find a set of 'n' elements {a₁, a₂, ..., a<0xE2><0x82><0x99>} such that:
13. G = ⋃ᵢ<0xE2><0x82><0x99>₁ aᵢH (where aᵢ are left coset representatives)
14. G = ⋃ᵢ<0xE2><0x82><0x99>₁ Haᵢ⁻¹ (where aᵢ⁻¹ are right coset representatives)
15. This proves that there exist elements (specifically, the inverses of the left coset representatives) that can serve as right coset representatives.