Set $\mathbb{K} = \mathbb{Q}((2)^{1/8},i)$. Prove that $Aut(\mathbb{K}/ \mathbb{Q}(\sqrt{-2})) = Q_8$.

Question

I first proved that $G = Gal(\mathbb{K} / \mathbb{Q}) = D_8$. Since, if we consider the following maps $\sigma_1(2^{1/8}) = (2)^{1/8} \mapsto (2)^{1/8}\omega$ and fixig i. Also, we consider $\theta_1(i) = -i$ and fixing the other root. I proved that those maps generate the whole galois group. I also proved that if $H = <\sigma_1>$ and $K = <\theta_1>$ then H is normal in G, $H \cap K = 1$, and $\theta_1 \sigma_1 \theta_1^{-1} = \sigma_1^3 $. This mean that $G = < a,b : a^8 = b^2 = 1, bab^{-1} = a^3>$. I noticed that this problem has an example in Dummit and foote, however their solution relies on figuring out the subgroup lattice, which I don't have an idea how they figured it out very fast. Can someone explain to my why it is obvious that the lattice is given by the image below ? If someone can explain this to me that would be nice.