If $m_1=m_2z$ and $n_1=n_2z$ where $z=\operatorname{lcm} (m_1,n_1)$, then $\operatorname{lcm}(m_2,n_2)=1$

Question

I know if $z=\operatorname{lcm}(m_1,n_1)$, then

(1) $n_1|z$ and $m_1|z$

(2) for every integer $k$, if $n_1|k$ and $m_1|k$, then $z|k$

and I know that $m_2|m_1$ and $n_2|n_1$ but I dont know what is the next?

Answer 1

(1).If $z=\operatorname{lcm}(m_1,n_1)$, then $m_1|z$ and $n_1|z$ by the definition of. $z$ is the common multiplier of $m_1$ and $n_1$.

(2).We prove by contradiction: If $k$ is not divided by $z$, then by division algorithm, $k=zd+r$ where $0<r<z$, by hypothesis that $m_1|k=zd+r$ and $n_1|k=zd+r$ and also $m_1|z$ and $n_1|z$, so $m_1|r$ and $n_1|r$, this lead to r is the common multiplier of $m_1$ and $n_1$and $r<z$, which contradict that z is the least common multiplier.