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


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

Is this direction right? I would appreciate any help.

Thanks and sorry if I have English mistakes

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

1 Answer 1


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

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


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

  • 1
    $\begingroup$ wow, thank you so much! $\endgroup$
    – DanielG
    Dec 2, 2021 at 10:55
  • $\begingroup$ it is true only for abelian group no? $\endgroup$
    – DanielG
    Dec 2, 2021 at 13:00
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 2, 2021 at 13:25
  • $\begingroup$ but how we assume $\left\langle x^{a},x^{b}\right\rangle $ is commutative? $\endgroup$
    – DanielG
    Dec 2, 2021 at 14:52
  • $\begingroup$ 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. $\endgroup$ Dec 2, 2021 at 15:18

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