Originally Posted by

**Drexel28** **NOTE: **I am actually not sure if this is correct. Let another member validate this. I feel as though I may have made a stupid incorrect assumption.

**Problem:** Let $\displaystyle (m,n)=1$. Prove that if $\displaystyle |a|=m,|b|=n$ that $\displaystyle |ab|=mn$.

**Proof:** Clearly $\displaystyle \left(ab\right)^{mn}=e$. Now suppose that $\displaystyle |ab|=\omega<mn$ then $\displaystyle (ab)^{\omega}=e\implies a^{\omega}=b^{-\omega}$ and $\displaystyle b^{\omega}=a^{-\omega}$. Thus $\displaystyle \left(b^{-\omega}\right)^m=\left(a^{\omega}\right)^m=e\impli es a^{\omega\cdot m-\omega\cdot n}=e$. Similarly $\displaystyle b^{\omega\cdot n-\omega\cdot m}=e$. And since $\displaystyle |a|=m$ we see that $\displaystyle m|\omega\cdot m-\omega\cdot n\implies m|\omega n$ and since $\displaystyle (m,n)=1$ this is only true if $\displaystyle m|\omega$. Similarly $\displaystyle n|\omega\cdot n-\omega\cdot m$ and since $\displaystyle |b|=n$ this means $\displaystyle n|\omega\cdot n-\omega\cdot m\implies n|\omega\cdot m\$ and by previous reasoning $\displaystyle n|\omega$. Therefore $\displaystyle m,n|\omega$ which means that $\displaystyle \text{lcm}(m,n)|\omega$ but since $\displaystyle (m,n)=1$ we have that $\displaystyle \text{lcm}(m,n)=mn$ and $\displaystyle mn|\omega$ which is a contradiction since $\displaystyle 0<\omega<mn$.