prove that the set H = 1 n 0 1 n exists in Z(integer) Is a cyclic subgroup of th

Question

prove that the set H = 1 n 0 1 n exists in Z(integer) Is a cyclic subgroup of the group of all invertible matrices in M2(R)

Explanation / Answer

[1 n] [1 m]....[1 (n+m)] [0 1] [0 1].=.[0.....1....] Effectively, the multiplication of matrices corresponds to adding n and m!! This suggests the map f : (Z, +) --> (H, *) where f(n) = [1 n] ........[0 1]. We have already essentially shown that f is a homomorphism: For integers m and n, f(n) f(m) = [1 n] [1 m]....[1 (n+m)] [0 1] [0 1].=.[0.....1....] = f(n + m). Next, f is surjective: For any matrix in H (with m in Z) [1 m] [0 1], note that f(m) equals this matrix. Finally f is injective, because its kernel equals {0}: ker(f) = {n in Z : f(n) = I} = {0}. So, f is an isomorphism. Finally, since (Z, +) is a cyclic group (with respect to addition), we may conclude that H is also cyclic.

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