Divisibility property of product two elements in abelian group

Question

Let $G$ be a finite abelian group and $d=o(ab), \ m=o(a), \ n=o(b)$.

Show that $d\mid \frac{mn}{\text{gcd}(m,n)}$ and $\frac{mn}{\text{gcd}(m,n)^2}\mid d$.

In particular, if $m$ and $n$ are coprime then order of product is multiplicative.

Proof: $(ab)^{\text{lcm}(m,n)}=a^{\text{lcm}(m,n)}b^{\text{lcm}(m,n)}=e$ then $d\mid \text{lcm}(m,n)$ or $d\mid \frac{mn}{\text{gcd}(m,n)}$. We have done with the first relation.

Since $e=(ab)^d=(ab)^{\text{gcd}(m,n)d}=a^{\text{gcd}(m,n)d}b^{\text{gcd}(m,n)d}$. If I'll show that $a^{\text{gcd}(m,n)d}=e$ and $b^{\text{gcd}(m,n)d}=e$ then $m\mid \text{gcd}(m,n)d$ and $n\mid \text{gcd}(m,n)d$ so we get what we need, i.e. $\frac{mn}{\text{gcd}(m,n)^2}\mid d$.

But as you see I have difficulties with showing that $a^{\text{gcd}(m,n)d}=e$.

Can anyone help with that, please?

Answer 1

If $ab$ has order $d$ in abelian group, then $(ab)^d=a^db^d=e$ so $a^d=b^{-d}$. Now write $(m, n)=ms+nt$ for some integers $s$ and $t$. So $$a^{(m, n)d}=a^{msd}\cdot a^{ntd}=b^{-ntd}=e.$$ Therefore $m|(m, n)d$. Similarly $n|(m, n)d$.