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Let $G$ and $H$ be groups, $g \in G$, and $h \in H$. Suppose that $\text{lcm}(|g|, |h|) = |G||H|$. I want to show that $|g| = |G|$ and $|h| = |H|$.

Do we use the fact that $|g| \leq |G|$ and $|h| \leq |H|$?

Thanks in advance.

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  • $\begingroup$ You need the slightly stronger fact that $|g|$ divides $|G|$ and $|h|$ divides $|H|$. As a start is $hcf(|g|,|h|)>1$ possible? $\endgroup$ – Robert Chamberlain Aug 18 at 20:23
  • $\begingroup$ The kind of $\LaTeX$ called MathJax works in the title section too, don't you know? $\endgroup$ – Shaun Aug 18 at 21:00
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Yes, we can. Since $|g| \le |G|$ and $|h| \le |H|$, we have $|g|\cdot |h| \le |G|\cdot |H|$ (since everything here is positive). We also know that $\text{lcm}(|g|,|h|) \le |g|\cdot |h|$. Therefore, $$\text{lcm}(|g|,|h|) = |G|\cdot |H| \implies |g|\cdot|h| = |G|\cdot |H|$$

Now Suppose for a contradiction that $|g| < |G|$. Then, in order for $|g|\cdot|h| = |G|\cdot |H|$ to hold, we need to have that $|h| > |H|$, which is a contradiction clearly. Since $|g| > |G|$ is not possible neither, we have $|g| = |G|$. Then $|h| = |H|$ also follows from the equation $|g|\cdot|h| = |G|\cdot |H|$.

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  • $\begingroup$ Thanks. That seems to work. $\endgroup$ – Tim Aug 18 at 21:26
  • $\begingroup$ You're welcome. Good luck! $\endgroup$ – ArsenBerk Aug 19 at 9:24
  • $\begingroup$ @Tim It's true for any integers with $\,g\mid G,\, h\mid H,\ {\rm lcm}(g,h) = GH,\,$ see my answer. $\endgroup$ – Bill Dubuque Aug 20 at 2:16
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For any integers: $\,h\mid H\,\Rightarrow\,g,h\mid gH\Rightarrow\, {\rm lcm}(g,h)\!=\!GH\mid gH\,\Rightarrow\,G\mid g,\,$ so $\,g\mid G\,\Rightarrow\, G = \pm g$

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  • $\begingroup$ $H = \pm h\,$ by symmetry. The proof works in any UFD or GCD domain, yielding $\,G,g\,$ and $\,H,h\,$ are associates. $\endgroup$ – Bill Dubuque Aug 20 at 2:10
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Write $|G|=a$, $|H|=b$, $|g|=x$ and $|h|=y$. Then $x\mid a$ and $y\mid b$. Write $a=px$ and $b=qy$, so $ab=pqxy$. By assumption, $\operatorname{lcm}(x,y)=ab$, so $xy=rab$, for some $r$ and we get $ab=pqrab$, so $pqr=1$, forcing $p=q=r=1$.

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  • $\begingroup$ Clearer to use the LCM universal property, e.g. see my answer. $\endgroup$ – Bill Dubuque Aug 20 at 2:05

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