Does the order of the subgroup generated by two elements divide the product of the element orders?

Question

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.

Groups have a wonderful amount of flexibility. You might be interested in the free product of groups. — Santana Afton, Apr 26 '19 at 07:03

score 4 · Answer 1 · answered Apr 26 '19 at 06:49

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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.

answered Apr 26 '19 at 06:49

Hagen von Eitzen

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score 3 · Accepted Answer · answered Apr 26 '19 at 06:32

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

answered Apr 26 '19 at 06:32

ancient mathematician

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Related: $S_n$ is generated by $(12)$ and $(123\cdots n)$, yet $n!$ rarely divides $2n$. – Greg Martin Apr 26 '19 at 08:45
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And for more fun https://mathoverflow.net/questions/59213/generating-finite-simple-groups-with-2-elements. – ancient mathematician Apr 26 '19 at 08:52

Does the order of the subgroup generated by two elements divide the product of the element orders?

2 Answers2