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

Question

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

Answer 1

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.

Answer 2

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.