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Assume in an abelian group $G$ that $\langle b\rangle\cap \langle a\rangle=e$, then the order of $(ab)$ is the lcm of the orders of $a$ and $b$. Essentially, $|ab|=\operatorname{lcm}(|a|,|b|)$.

So far I have the assumption, then suppose $(ab)^k=e$, where $k$ is the smallest integer. So $|ab|=k$. Since $G$ is abelian, $(ab)^k=a^k b^k=e$. Thus, $(a^k)^{-1}=b^k$ and likewise the same goes for, $(b^k)^{-1}=a^k$. Therefore $(b^k)^{-1}\in\langle a\rangle$ and $(a^k)^{-1}\in\langle b\rangle$. So each of $a^k$ and $b^k$ are the identity. Therefore $k$ is a multiple of $n$ and $m$, so we can say $\operatorname{lcm}(m,n)\mid k$. How do I get the last piece I need?

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  • $\begingroup$ Just so you know, you don't have to have the condition that $\langle a \rangle \cap \langle b \rangle$ contains the identity, since both of those are subgroups of $G$, and all subgroups of $G$ must contain $e$ otherwise they don't preserve the 'group' structure. I think you have to assume that $a$ and $b$ have finite order, and that $a$ is not a power of $b$ and vice versa. $\endgroup$ – Bruce Zheng Oct 9 '14 at 20:14
  • $\begingroup$ yes I know, but that is how the question starts off. $\endgroup$ – Jack Armstrong Oct 9 '14 at 20:14
  • $\begingroup$ Gotcha. Guess the question was written a little strangely, then. $\endgroup$ – Bruce Zheng Oct 9 '14 at 20:15
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Let $n = \text{lcm}(|a|,|b|)$. So we can write $n = |a|r = |b|s$ for some positive integers $r$ and $s$.

Then $(ab)^{n} = a^{n}b^{n} = (a^{|a|})^{r}(b^{|b|})^{s} = e^{r}e^{s} = e$, so $|ab|$ divides $n$.

On the other hand, your argument above shows that $n$ divides $|ab|$, and so we have equality.

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($k\neq 0$).

This show that $\mathrm{lcm}(|a|,|b|)\leq k$.

$(ab)^{\mathrm{lcm}(|a|,|b|)}=e$ , hence $k=\mathrm{lcm}(|a|,|b|)$ ($k$ is the smallest integer $\neq 0$).

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  • $\begingroup$ how is this true? $(ab)^{\mathrm{lcm}(|a|,|b|)}=e$ $\endgroup$ – Jack Armstrong Oct 9 '14 at 20:10
  • $\begingroup$ @Jack $(ab)^r=a^rb^r$ $\endgroup$ – egreg Oct 9 '14 at 20:15
  • $\begingroup$ With the comment of @egreg , and that $a^{\mathrm{lcm}(|a|,|b|)}=e$ and $b^{\mathrm{lcm}(|a|,|b|)}=e$ $\endgroup$ – Hamou Oct 9 '14 at 20:17
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    $\begingroup$ $(ab)^{\mathrm{lcm}(|a|,|b|)}=e$ implies that $k\leq \mathrm{lcm}(|a|,|b|)$, because $k$ is the smallest with this property. But we have also $ \mathrm{lcm}(|a|,|b|)\le k$, hence $k=\mathrm{lcm}(|a|,|b|)$. $\endgroup$ – Hamou Oct 9 '14 at 20:25
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    $\begingroup$ $\mathrm{lcm}(|a|,|b|)$ is a multiple of $|a|$ hence $a^{\mathrm{lcm}(|a|,|b|)}=e$,and $\mathrm{lcm}(|a|,|b|)$ is a miltiple of $|b|$ hence $b^{\mathrm{lcm}(|a|,|b|)}=e$, and $(ab)^{\mathrm{lcm}(|a|,|b|)}=a^{\mathrm{lcm}(|a|,|b|)}b^{\mathrm{lcm}(|a|,|b|)}=ee=e$ $\endgroup$ – Hamou Oct 9 '14 at 20:31

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