# Does the order of the subgroup generated by two elements divide the product of the element orders? [duplicate]

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 $$\left|\left\right|$$ 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.

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

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• The order might be infinite. – Trebor Apr 26 at 6:28
• Groups have a wonderful amount of flexibility. You might be interested in the free product of groups. – Santana Afton Apr 26 at 7:03

## 2 Answers

$$S_3=\langle (12), (23)\rangle$$.

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.