Assume $(G,\times)$ is a group and for $a,b \in G$: $ab=ba$, $\text{ord}(a)=n$, $\text{ord} (b)=m$ Assume $(G,\times)$ is a group and for $a,b \in G$: $ab=ba$, $\text{ord}(a)=n$, $\text{ord}
(b)=m$

*

*Show that if $\gcd(m,n)=1$ then $G$ has an element of order $nm$.

*If $m,n$ are arbitrary,then $G$ has an element of order $\text{lcm}(m,n)$

Since $G$ is not cyclic I don't have any idea how to start, any help is appreciated.

Lemma: Assume $(G,\times)$ is a group and $a,b \in G$, Moreover $ab=ba$.
let $\text{ord}(a)=n$ and $\text{ord}(b)=m$,then $\text{ord}(ab)\mid \text{lcm}(n,m)$.
$\text{lcm}(n,m)=ns$ and $\text{lcm}(n,m)=mr$ for some $r,s \in \mathbb Z^+$,then:
$$(ab)^{\text{lcm}(n,m)}$$ Since $ab=ba$ ,hence $$=a^{\text{lcm}(n,m)}b^{\text{lcm}(n,m)}$$
$$=a^{ns}b^{mr}=(a^n)^s(b^m)^r$$
$$=e^se^r=e$$
Follows $\text{ord}(ab)\mid \text{lcm}(n,m)$.


*

*Since $\text{ord}(ab) \mid \text{lcm}(n,m)=\frac{nm}{\text{gcd}(n,m)}$,By the assumption $\text{gcd}(n,m)=1$ So $\text{ord}(ab) \mid nm$

*If $\text{ord}(ab)
 \mid \text{lcm}(n,m)$ then there is $g \in G$ such that $g^{\text{lcm}(n,m)}=e$

For the first question, this trick can help you : if $(ab)^k=a^kb^k=1$ therefore $a^k = b^{-k}$ so $a^{mk}=b^{-mk}=1$ so $n \mid mk$ and since $\gcd(m,n)=1$ we have $n \mid k$ and $m \mid k$ (by symmetry) so $k= \alpha n= \beta m$ so using the same argument we have $nm \mid k$.
For the second one, remember that $\text{lcm}(m,n)=\frac{mn}{\gcd(m,n)}$ so if $m'=\frac{m}{pgcd(m,n)}$ and $n'=\frac{n}{pgcd(m,n)}$ we have
$$
m'n'=\text{lcm}(m,n) \text{ and } \gcd(m',n')=1,
$$
we get the result using $(1)$.
