# $\text{lcm}(|g|,|h|) = |G||H|$ implies $|g| = |G|$ and $|h| = |H|$

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 $$|h| \leq |H|$$?

• You need the slightly stronger fact that $|g|$ divides $|G|$ and $|h|$ divides $|H|$. As a start is $hcf(|g|,|h|)>1$ possible? – Robert Chamberlain Aug 18 at 20:23
• The kind of $\LaTeX$ called MathJax works in the title section too, don't you know? – Shaun Aug 18 at 21:00

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 $$|g|\cdot|h| = |G|\cdot |H|$$.

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

For any integers: $$\,h\mid H\,\Rightarrow\,g,h\mid gH\Rightarrow\, {\rm lcm}(g,h)\!=\!GH\mid gH\,\Rightarrow\,G\mid g,\,$$ so $$\,g\mid G\,\Rightarrow\, G = \pm g$$

• $H = \pm h\,$ by symmetry. The proof works in any UFD or GCD domain, yielding $\,G,g\,$ and $\,H,h\,$ are associates. – Bill Dubuque Aug 20 at 2:10

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 $$p=q=r=1$$.

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