$a=mx$ and $b=my$, $\gcd(x,y)=1$ and $p,q$ relatively prime to $a,b$. Is $\mu q=\tau q\pmod m$ for $\mu=p^{-1} \in Z_a$ and $\tau = p^{-1}\in Z_b$ Let $a=mx$ and $b=my$ where $\gcd(x,y)=1$. Let $p,q$ be integers relatively prime to $a$ and $b$. Let $\mu$ be the integer which represents the class of the multiplicative inverse of $p$ in $\mathbb{Z}_a$ and, similarly, let $\tau$ be the multiplicative inverse of $p$ in $\mathbb{Z}_b$.
Is it always the case that $\mu q =\tau q$ (mod m).
Every example with specific numbers that I have worked out under these conditions this has been the case, but I'm not sure how to prove it in general.

Yes it is. As $q$ is invertible $\mod{m}$, you are essentially asking if $\mu \equiv \tau \mod{m}$.
As $\mu p \equiv 1 \mod{a}$, $\mu p=ka +1=kmx+1$ for some $k \in \mathbb{Z}$, and so $\mu p \equiv 1 \mod{m}$.
A similar reasoning with $\tau$ gives us that $\tau p \equiv 1 \mod{m}$. We deduce that $\tau p \equiv \mu p \mod{m}$. Now, $p$ is invertible$\mod{m}$, and so this implies that $\mu \equiv \tau \mod{m}$.
