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First of all I'm new to group theory.

  1. Let $G$ be a group with elements $a$ and $b$ of finite order which commute. Suppose that for each $m \in \mathbb{Z}$, $a^m$ is not a power of $b$, and that for each $k \in \mathbb{Z}$, $b^k$ is not a power of $a$.

For example, this happens if $a$ and $b$ are disjoint permutations in $S_n$. Observe also that this ensures that, if $(a^n) (b^n) = e$, then $a = e$ and $b = e$.

Show that the order of ab is the least common multiple of the order of $a$ and the order of $b$

I tried proving this but I dont seem to be getting anywhere closer to a complete proof. I keep ending up in dead ends. Any help on the topic would be greatly appreciated.

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    $\begingroup$ Show us what you tried and where you think the dead end is. Learn to format the mathematics with mathjax - the ? icon on the right of the edit menu will help. (I did a little for you.) $\endgroup$ – Ethan Bolker Mar 12 '17 at 19:09
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What you really mean is that when $k \neq 0$ $a^k$ is not a power of $b$ and when $k \neq 0$ $b^k$ is not a power of $a$.

Lemma take $a, b \in G$ of finite order and which commute. Suppose $a^k$ is not a power of $b$ when $k \neq 0$ and $b^k$ is not a power of $a$ when $k \neq 0$. Then $ab$ has finite order and $|ab| = lcm(|a|, |b|)$, where $|g|$ is the order of an element in $G$.

Suppose $(ab)^k = 0$. Then $a^k = b^{|b| - k}$. Since $a^k$ is not a power of $b$ when $k \neq 0$, we must have $|a|$ divides $k$. Similarly, $|b|$ divides $k$. Therefore $lcm(|a|, |b|)$ divides $k$.

Conversely, $(ab)^{lcm(|a|, |b|)} = a^{lcm(|a|, |b|)} b^{lcm(|a|, |b|)} = e$ since $|a|$ and $|b|$ both divide $lcm(|a|, |b|)$.

Therefore $lcm(|a|, |b|)$ is the smallest positive number in $\{k \in \mathbb{Z} : (ab)^k = e \}$. So $|ab| = lcm(|a|, |b|)$.

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