$\operatorname{ord}(u)=r,\,\operatorname{ord}(v)=s$ $\,\Rightarrow\,\operatorname{ord}(uv)=rs\,$ if $\,r,s\,$ (co)primes

Let u and v be elements of a commutative group, and suppose that their orders are r and s, respectively. Show that if r and s are distinct primes then the order of uv is rs.

I will use the follwing theorem to proof this.

Theorem: Let x be an element of order m in a finite group G. Then $$x^s=1$$ in G if and only if s i a multiple of m.

My attempt

We can here see that $$(uv)^{rs}=[commutative]=u^{rs}v^{rs}=1^s1^r=1$$ By the theorem, this tells us that this is true only if only rs is a multiple of the order of $uv$. Let the order be $ord(uv)=k$. We know that every positive integer can be written as $$a=bm+l, 0<=l<b$$ and since $r$ and $s$ is primes (positive integers), $rs$ can be written as $$rs=km+l, 0<=l<k$$ This gives ut that $$1=(uv)^{rs}=(uv)^{km+l}=1^muv^l=(uv)^l$$ If l>0 then the equation $(uv)^l=1$ contradicts that the positive integer k is the least positive integer for which $(uv)^k=1$. Therefore l=0 and $rs=km$ . We can here see that $rs$ is a multiple of k (and this proof also the theorem). $$rs=km => k | rs => k|r ORk|s$$ But we know that $r$ and $s$ are distinct primes, so k must be equal to one of the primes.

$$k=r => (uv)^k=(uv)^r=1v^r=v^r=1$$ And since we know that $ord(v)=s$ the theorem tells us that $r$ must be a multiple of $s$. But since both of them is primes, we get that $r=s$ but this is a contradiction (r and s are distincts as the exercise tells us).

Now, we try with $k=s$. Same as above, we get that $s=r$ which is a contradiction. Neither k=r nor k=s is right since by the theorem we get that $k=r=s$ and we have that $r$ and $s$ are distincts. This tells us that k cannot be equal to r nor s, since this gives us that $r=s$. And since we know (and as i did proof)that $$rs=km$$, this tells us, as i wrote before, that $$k|rs$$ Since k is neither equal to $r$ or $s$, this tells us that k must be equal to $rs$. Therefore $$ord(uv)=k=rs$$.

Edit: i know that this can be prooved by l.c.m(x,y) but we have not used it so far. That's why i did not proof it with l.c.m(x,y)

There is a gap: the prime product $\,rs\,$ has three smaller divisors $\,r,s,\color{#c00}1.\,$ You've ruled out the possibilities $\, k = r,s\,$ but not $\,k = \color{#c00}1.\ $ Below is a quicker way to proceed, where we prove it more generally for any $r,s$ that are coprime, i.e. their gcd $(r,s) = 1.\,$ Let $\,o(x) := {\rm ord}\,x$

$ (uv)^k\! = 1\,\Rightarrow\, u^k\! = v^{-k} =\color{#c00}t \in \langle u\rangle\cap\langle v\rangle\Rightarrow \bigg\lbrace\begin{eqnarray} && {o(t)\mid\ r,s = o(u),\,o(v)},{\rm\ by\ Lagrange}\\ \Rightarrow && o(t)\mid(r,s)\!=\!1\, \Rightarrow\, \color{#c00}{t = 1}\end{eqnarray}$

So $\ (uv)^k\! = 1\iff u^k =\color{#c00}{1} = v^k\iff r,s\mid k\underset{\large (r,s)\,=\,1}\iff rs\mid k,\, $ so $\ rs = o(uv)\ $ by here.

Remark $ $ If Lagrange in unknown, instead: $\,t^{\large r}\! = (u^{\large k})^{\large r}\! = (u^{\large r})^{\large k}\! = 1^{\large k}\! = 1$ thus $\,o(t)\mid r$

To prove that (co)primes $\,r,s\mid k\,\Rightarrow\,rs\mid k\,$ without knowledge of lcm, you can either use the Fundamental Theorem of Arithmetic, or equivalent properties such as Euclid's Lemma, e.g. $\, s\mid k\,\Rightarrow\, k = sn,\,$ so $\,(r,s)=1,\,r\mid sn\,\Rightarrow\, r\mid n,\,$ so $\,n = rm,\,$ so $\,k = sn = s(rm),\,$ so $\,rs\mid k.$ Alternatively, using Bezout's identity, since $\,s,r\,$ are coprime there are integers $\,j,k\,$ with $\,js\!+\!kr = 1\,$ so $\,r\mid sn,rn\,\Rightarrow\,r\mid j(sn)+k(rn) = (js\!+\!kr)n = n,\,$ so the result follows as above. See also this answer for further proofs and comparisons.

One can use the same (correct) idea, streamline things a bit, and relax the primality assumptions.

Suppose that $u$ has order $r$ and $v$ has order $s$, where $r$ and $s$ are relatively prime. We show that $uv$ has order $rs$. Since $(uv)^{rs}=1$, it is enough to show that if $k$ is the order of $uv$, then $rs$ divides $k$.

Note that $u^{ks}=u^{ks}v^{ks}=1$. So the order $r$ of $u$ divides $ks$, and since $r$ and $s$ are relatively prime, we conclude that $r$ divides $k$.

Similarly, $s$ divides $k$.

Since $r$ and $s$ are relatively prime, it follows that $rs$ divides $k$.