Let G denotes the group of all 2 x 2 invertible matrices with entries ...
To show that H1 is a subgroup of G, we need to show that it satisfies the three conditions for a subgroup:
1. Closure: For any two matrices A and B in H1, their product AB must also be in H1.
Consider two matrices A and B in H1, such that A = [[a, b], [c, d]] and B = [[e, f], [g, h]]. We have:
AB = [[a, b], [c, d]] [[e, f], [g, h]]
= [[ae + bg, af + bh], [ce + dg, cf + dh]]
Since A and B are invertible, their determinants are non-zero, so ad - bc ≠ 0 and eh - fg ≠ 0. Therefore, the determinant of AB is given by:
det(AB) = (ae + bg)(cf + dh) - (af + bh)(ce + dg)
= aecf + aedh + bccf + bcdh - aecf - aedg - bcfg - bcdh
= aedh - aedg - bcfg
= ad(eh - fg)
Since ad - bc ≠ 0, we have det(AB) ≠ 0, which means AB is invertible. Hence, AB ∈ H1, and closure is satisfied.
2. Identity: The identity matrix I = [[1, 0], [0, 1]] is in H1, since its determinant is 1.
3. Inverse: For any matrix A in H1, its inverse A^(-1) = [[d, -b], [-c, a]] is also in H1, since its determinant is (ad - bc)^(-1), which is non-zero.
Therefore, H1 is a subgroup of G.