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I was just hoping to get a quick verification of my proof for the following (common) assertion:

If $\langle x\rangle \cap\langle y\rangle=e$ for group elements in a group $G$ that commute (i.e. $xy=yx$), then $|xy|=\text{lcm}(|x|,|y|)= [|x|,|y|]$.

Proof:

Let $c=|xy|$ and $d=[|x|,|y|]$. Then the division algorithm furnishes a $q\in\mathbb{Z^+}$ and $0\leq r< d$ such that $c = dq+r$. Then $$(xy)^c=e=(xy)^{dq+r}=\left((xy)^d\right)^q(xy)^r$$

Since the orders of $x$ and $y$ divide $d$ and $xy=yx$, $(xy)^d=x^dy^d=e\cdot e= e$ so that $\left((xy)^d\right)^q=e^q=e.$

Thus $e=e\cdot(xy)^r=(xy)^r=x^ry^r \iff x^r=y^{-r}$. However, the last equation implies for nonzero $r$ that powers of $x$ are contained in $\langle y\rangle$ and powers of $y$ are contained in $\langle x \rangle$, which contradicts that we have $\langle x\rangle \cap\langle y\rangle=e$. Thus, we must have $r=0$, so that $c=dq$. We also know that in general for any commuting elements in a group, $cq^*=d$ for some $q^*\in\mathbb{Z^+}.$ Thus combining, we have that $|xy|= [|x|,|y|]$.

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  • $\begingroup$ May I know what is meant by your notation $[|x|,|y|]$? $\endgroup$ Jul 12, 2022 at 7:00
  • $\begingroup$ Of course. I just mean the least common multiple of the numbers $|x|$ and $|y|$ by the notation $[|x|,|y|]$. $\endgroup$
    – user689775
    Jul 12, 2022 at 7:04

2 Answers 2

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Some comments on your proof:
First of all, the question should assume that $|x|$ and $|y|$ are both finite.

Next, the proof for $d$ divides $c$ is fine. There is a little gap on the statement "We also know that in general for any commuting elements in a group, $cq^*=d$ for some $q^*\in\mathbb{Z^+}.$" Since $d$ is a multiple of $|x|$ and $|y|$, we have $x^d=y^d=e$. Therefore we have $(xy)^d=x^dy^d=ee=e$ and conclude that $c=|xy|$ divides $d$. At last, we obtain $c=d$ because they are positive integers that divide each other.

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  • $\begingroup$ Thanks, Alan! I've been thinking about the lack of the finiteness assumption as well. I was thinking that the question (Exercise $1.5$, from Larry Grove's Algebra) implied as much by it merely using the symbols $|x|$ and $|y|$: "If $x$ and $y$ are commuting elements in a group $G$, show that $|xy|$ divides $\text{LCM}(|x|,|y|)$; equality holds if $\langle x \rangle \cap \langle y \rangle = 1.$" $\endgroup$
    – user689775
    Jul 12, 2022 at 7:15
  • $\begingroup$ @upanddownintegrate You are welcomed. By the way this is the first time I see the notation $[a,b]$ that denote $\text{lcm}(a,b)$. $\endgroup$ Jul 12, 2022 at 8:54
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Take the mapping $\langle x\rangle \times \langle y\rangle \rightarrow \langle x,y\rangle:(x^i,y^j)\mapsto x^iy^j$. This is a surjective group homomorphism (as xy=yx) with trivial kernel by hypothesis. By the first isomorphism theorem, preimage and image have the same size.

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