Proof groups equality: $\langle x^{t_1},x^{t_2} \rangle = \langle x^{\gcd(t_1,t_2,n)} \rangle$

Question

Let $C_n$ be a cyclic group of order n. I need to prove $<x^{t_1},x^{t_2}> = <x^{\gcd(t_1,t_2,n)}>$ where gcd is the greatest common divisor of $t_1,t_2,n$ and $<>$ denotes the subgroup generated by the elements between the brackets.

It's easy to proof that $<x^{t_1},x^{t_2}> \subseteq <x^{\gcd(t_1,t_2,n)}>$ but I don't know how to do the other inclusion.

Answer 1

Divide the proof into two simpler ones:

• $\langle x^{t_1},x^{t_2} \rangle = \langle x^{\gcd(t_1,t_2)} \rangle$

• $\langle x^{t} \rangle = \langle x^{\gcd(t,n)} \rangle$

Both proofs are based on Bézout's identity.