I'm trying to generalize my previous lemma to this well-known result. Could you please confirm if it is fine or contains logical mistakes?
Let $G$ be an abelian group with elements $x, y$ of orders $m$ and $n$ respectively. There exists $z \in G$ of order $\operatorname{lcm} (m,n)$.
My attempt:
My lemma is
Let $G$ be an abelian group with elements $x, y$ of orders $m$ and $n$ respectively. Moreover, $\gcd(m,n)=1$. Then the order of $xy$ is $mn$.
Let $d = \gcd(m,n)$ and $p = m/d$ and $q =n/d$. Then $\gcd(p,q) = 1$ and $o(x^d) =p$ and $o(y^d) =q$. Apply the lemma to $x^d$ and $y^d$, we get $o(x^d y^d) = pq$. We again have $\gcd(pq, d) = 1$ and $o \left (x^{p} \right ) = d$. Applying the lemma again to $x^d y^d$ and $x^p$, we get $o(x^d y^dx^p) = pqd = \operatorname{lcm} (m,n)$.
Update: I've just found a fix
We factorize $\operatorname{lcm} (m,n) = \prod p_i^{r_i}$. Then for each $i$, we have $p_i^{r_i}$ divides at least one of two numbers $m$ and $n$. We pick $h_i$ from the set $\{h \in \{m, n\} | o(h) \text{ is divisible by } p_i^{r_i} \}$ and $k_i = o(h_i) / p_i^{r_i}$.
We have $\gcd(p_1^{r_1} , p_2^{r_2} ) = 1$ and $o( h_1^{k_1}) = p_1^{r_1}$ and $o( h_2^{k_2}) = p_2^{r_2}$. By our lemma, $o(h_1^{k_1} h_2^{k_2}) = p_1^{r_1} p_2^{r_2}$. Continue this process, we get $o(\prod h_i^{k_i}) = \operatorname{lcm} (m,n)$.
