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Q: Suppose that $a,b$ are elements of finite order in a group such that $ab=ba$ and $\langle a\rangle\cap\langle b\rangle =\{e\}$. Prove that $o(ab)=\operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle))$

Attempt:

1) Showing that $o(ab)\mid \operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle))$

Let $o(\langle a\rangle)=m,o(\langle b\rangle)=n$

Then $\operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle))=mk_1=nk_2,\quad for \;some\quad k_1,k_2\in\mathbb Z $

$(ab)^{\operatorname{lcm}(o(\langle a\rangle ),o(\langle b\rangle))}=a^{\operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle))}b^{\operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle>))}$

That is $(ab)^{\operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle))}=a^{mk_1}b^{nk_2}=e$

Then $o(ab)\mid\operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle))$

2) I am stuck here how can I show $\operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle))\mid o(ab)$

Is this method good, if it is how I should start 2. part?

What others method we can apply?

Note: By the way, I fully understand the theorem but I feel like I need to prove this precisely.

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It is a fact that $\operatorname{lcm}(x,y)$ is the least common multiple of $x$ and $y$ both with respect to $<$ and to the divisibility relation $|$. Therefore, in order to show that $\operatorname{lcm}(o(\langle a\rangle),o(\langle b\rangle))|o(ab)$, it is enough to prove that $o(ab)$ is a common multiple of $o(\langle a\rangle)$ and of $o(\langle b\rangle)$. For this, we only have to prove that $a^{o(ab)}=b^{o(ab)}=e$.

We know that $(ab)^{o(ab)}=a^{o(ab)}b^{o(ab)}=e$ hence $a^{o(ab)}=b^{-o(ab)}\in <a>\cap<b>$. It follows by hypothesis that $a^{o(ab)}=b^{o(ab)}=e$.

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  • $\begingroup$ I cannot believe that I missed :( $\endgroup$ – Azer BABA fanlari Jun 13 '18 at 13:06

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