Let $\varphi: { GL_{2}(\mathbb{Z}_{7}) \to U_{7} } $ homomorphism such that $\varphi(A) = det(A) $.
Proof that $H = GL_{2}(\mathbb{Z}_{7}) / ker(\varphi)$ is cyclic and find its generator.
My try: from first homomorphism theorem we know that $O(H) = 6$ because $\varphi$ is on $U_{7}$, but I'm not sure how to continue...
How can I prove that the quotient group is cyclic before finding its generator?
By the first isomorphism theorem $\operatorname{GL}_2(\Bbb{Z}/7\Bbb{Z})/\ker\varphi$ is isomorphic to a subgroup of $U_7$. So it suffices to show that $U_7$ is cyclic. The easiest way to prove this depends on your definition of $U_7$.