Problem statement: Construct a nonabelian group of order 44
The answer claims that $Z_{11} \rtimes (Z_2 \times Z_2) \cong \langle a,b,c|a^{11}=1, b^2=c^2=1, bc=cb, b^{-1}ab = c^{-1}ac = a^{-1} \rangle$. But since only one of the element from $Z_2 \times Z_2$ is mapped to the inverse automorphism, shouldn't it be either $b^{-1}ab=a^{-1}$ OR $c^{-1}ac=a^{-1}$?
No. This is just one way of naming the generators. Note that $(bc)^{-1}a(bc)=a$, so you can replace $c$ by $bc$ to get your desired presentation.