# If $\langle x\rangle \cap\langle y\rangle=e$, then $|xy|=[|x|,|y|]$.

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 $$|xy|= [|x|,|y|]$$.

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

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.
• 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 $\langle x \rangle \cap \langle y \rangle = 1.$"
• @upanddownintegrate You are welcomed. By the way this is the first time I see the notation $[a,b]$ that denote $\text{lcm}(a,b)$. Jul 12, 2022 at 8:54
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.