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Let $(G, \cdot)$ be a group, $a,b\in G$ such that $\DeclareMathOperator{\ord}{ord}\ord(a),\ord(b)<\infty$.

Do we then have that $\left|\left<a,b\right>\right|$ divides $\ord(a)\ord(b)$? (Where $\left< a,b \right>$ denotes the subgroup generated by $a$ and $b$)

I managed to show this for the case that $a$ and $b$ commute, however I would be interested if this holds even if they don't.

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marked as duplicate by Asaf Karagila Apr 26 at 9:48

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  • $\begingroup$ The order might be infinite. $\endgroup$ – Trebor Apr 26 at 6:28
  • $\begingroup$ Groups have a wonderful amount of flexibility. You might be interested in the free product of groups. $\endgroup$ – Santana Afton Apr 26 at 7:03
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$S_3=\langle (12), (23)\rangle$.

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The group $$\left\langle\begin{pmatrix}0&-1\\1&0\end{pmatrix},\begin{pmatrix}0&-1\\1&1\end{pmatrix}\right\rangle$$ has infinite order, in spite of the generators having finite order.

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