  Let $M(x)=\sum_{1\le n\le x}\mu(n)$ where $\mu$ is the M\"obius function. It
is well-known that the Riemann Hypothesis is equivalent to the assertion that
$M(x)=O(x^{1/2+\epsilon})$ for all $\epsilon>0$. There has been much interest
and progress in further bounding $M(x)$ under the assumption of the Riemann
Hypothesis. In 2009, Soundararajan established the current best bound of \[
M(x)\ll\sqrt{x}\exp\left((\log x)^{1/2}(\log\log x)^c\right) \] (setting $c$ to
$14$, though this can be reduced). Halupczok and Suger recently applied
Soundararajan's method to bound more general sums of the M\"obius function over
arithmetic progressions, of the form \[ M(x;q,a)=\sum_{\substack{n\le x \\
n\equiv a\pmod{q}}}\mu(n). \] They were able to show that assuming the
Generalized Riemann Hypothesis, $M(x;q,a)$ satisfies \[
M(x;q,a)\ll_{\epsilon}\sqrt{x}\exp\left((\log x)^{3/5}(\log\log
x)^{16/5+\epsilon}\right) \] for all $q\le\exp\left(\frac{\log 2}2\lfloor(\log
x)^{3/5}(\log\log x)^{11/5}\rfloor\right)$, with $a$ such that $(a,q)=1$, and
$\epsilon>0$.
  In this paper, we improve Halupczok and Suger's work to obtain the same bound
for $M(x;q,a)$ as Soundararajan's bound for $M(x)$ (with a $1/2$ in the
exponent of $\log x$), with no size or divisibility restriction on the modulus
$q$ and residue $a$.
