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

Question

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 $x^d=x^{ma}\cdot x^{nd}$. — stephenkk, Dec 02 '21 at 10:48

score 2 · Accepted Answer · answered Dec 02 '21 at 10:46

2

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.

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

answered Dec 02 '21 at 10:46

TheSilverDoe

29,720

1

wow, thank you so much! – DanielG Dec 02 '21 at 10:55
it is true only for abelian group no? – DanielG Dec 02 '21 at 13:00
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@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. – TheSilverDoe Dec 02 '21 at 13:25
but how we assume $\left\langle x^{a},x^{b}\right\rangle $ is commutative? – DanielG Dec 02 '21 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. – TheSilverDoe Dec 02 '21 at 15:18
Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Dec 19 '21 at 03:07

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

1 Answers1