I had to solve an exercise comprised of 3 parts:
a) For q, a integers, $q\geq 2$ and $\gcd(a,q)=1$ to show that $$\lim_{x \rightarrow \infty} \frac{1}{x}\mkern-18mu \sum_{\substack{n \leq x\\n \equiv a\pmod q}}\mkern-18mu \mu (n) = 0$$
b) For $m \in \mathbb{Z}_{>0}$, to show
$$\lim_{x \rightarrow \infty} \frac{1}{x}\mkern-18mu \sum_{\substack{n \leq x\\n \equiv a\pmod q ,\\ \gcd(m,n)=1}}\mkern-18mu\mu (n) = 0$$
c) For $q, a$ integers, $q\geq 2$ and $\gcd(a,q)=d>1$, to show that $$\lim_{x \rightarrow \infty} \frac{1}{x}\mkern-18mu \sum_{\substack{n \leq x\\n \equiv \text{a(mod q)}}}\mkern-18mu \mu (n) = 0$$
Out of this I solved a), b) using ideas from Dirichlet characters, Tauberian Theorem and some other tools from complex analysis, but for c) I cannot seem to get anywhere. As a small hint, I need to embed b) but not sure how this would work. I would appreciate any suggestion in the right direction.
If $\gcd(a,q)=d$ then the change of variables $n=kd$ yields $$ \sum_{\substack{n\le x \\ n\equiv a\pmod q}} \mu(n) = \sum_{\substack{k\le x/d \\ k\equiv a/d\pmod{n/d}}} \mu(kd) = \mu(d) \sum_{\substack{k\le x/d \\ k\equiv a/d\pmod{n/d} \\ (k,d)=1}} \mu(k). $$