Final answer:
In a finite group G, if H is a subgroup of index 2, there exist only two cosets: H and aH. Due to the limited number of cosets and the properties of a subgroup, every left coset aligns with a right coset. This is a consequence of the index of H in G being 2 and the partitioning of the group by its cosets.
Step-by-step explanation:
The question pertains to group theory, a fundamental area of abstract algebra. It asks us to demonstrate why, in a finite group G where H is a subgroup of index 2, every left coset of H is also a right coset of H. The index of a subgroup is the number of left cosets (or, equivalently, right cosets) that the subgroup has in the group, which, in this case, is 2.
Let's denote the two distinct cosets of H in G as H itself (since every subgroup is a coset of itself) and aH for some a in G not in H. We know that G is the disjoint union of these cosets, so any element g in G belongs to either H or aH. The key insight for this proof is that cosets partition the group, and in the case of index 2, there's only one possible partition other than the subgroup itself. If g is in H, then the right coset Hg equals H, and we have both a matching left and right coset. If g is in aH, then for some h in H, g = ah. The right coset Hg is then H(ah), which equals (Ha)h because H is a subgroup (H is closed under group operations and has associativity), given as Ha since a is not in H. As there are only two cosets, Ha must be aH, since H is the other coset and a cannot be in H. Hence, we see that every left coset is also a right coset, as required.