Prove: $\forall a,b\in\mathbb{Z}$, $x\in G$ : $\left\langle x^{a},x^{b}\right\rangle =\left\langle x^{d}\right\rangle$ [duplicate]

Let $$G$$ be a group, and let $$d=\gcd(a,b)$$

Prove: $$\forall a,b\in\mathbb{Z}$$, $$x\in G$$:

$$\left\langle x^{a},x^{b}\right\rangle =\left\langle x^{d}\right\rangle.$$

My attempt was to first prove that: $$\left\langle a,b\right\rangle =\left\langle d\right\rangle$$

and then maybe rely on this proof, to show that $$\left\langle x^{a},x^{b}\right\rangle \subseteq\left\langle x^{d}\right\rangle$$

and

$$\left\langle x^{a},x^{b}\right\rangle \supseteq\left\langle x^{d}\right\rangle$$

Is this direction right? I would appreciate any help.

Thanks and sorry if I have English mistakes

• d=gcd(a,b), thus there exists integers m,n, such that d=ma+nd. So $x^d=x^{ma}\cdot x^{nd}$. Dec 2, 2021 at 10:48

Because $$x^a$$ and $$x^b$$ commute, then $$\langle x^a, x^b \rangle = \lbrace (x^a)^n(x^b)^m, n, m \in \mathbb{Z} \rbrace$$

so $$\langle x^a, x^b \rangle = \lbrace x^{na+mb}, n, m \in \mathbb{Z} \rbrace$$

Now, $$\lbrace na+mb, n,m \in \mathbb{Z} \rbrace = \lbrace kd, k \in \mathbb{Z}\rbrace$$ by definition of $$d$$ as the generator of $$\langle a,b \rangle$$ in $$\mathbb{Z}$$. So

$$\langle x^a, x^b \rangle = \lbrace x^{kd}, k \in \mathbb{Z} \rbrace = \lbrace (x^{d})^k, k \in \mathbb{Z} \rbrace$$

i.e.

$$\boxed{\langle x^a, x^b \rangle = \langle x^d \rangle}$$

• wow, thank you so much! Dec 2, 2021 at 10:55
• it is true only for abelian group no? Dec 2, 2021 at 13:00
• @DanielG The key point is indeed that $x^a$ and $x^b$ commute, but this is always true, even if the whole group is not abelian. Dec 2, 2021 at 13:25
• but how we assume $\left\langle x^{a},x^{b}\right\rangle$ is commutative? Dec 2, 2021 at 14:52
• Because $x^ax^b = x^{a+b} = x^{b+a} = x^b x^a$. So $x^a$ and $x^b$ always commute (even if $G$ is not abelian), and this implies directly that $\langle x^a, x^b \rangle$ is abelian. Dec 2, 2021 at 15:18