For a homework exercise, I'm to determine for each $p$ the number of non-isomorphic tamely ramified Galois extensions $K/\mathbb{Q}_p$ such that $\operatorname{Gal}(K/\mathbb{Q}_p) \cong \mathcal{Q}_8$ (the quaternion group of order $8$).
I was out of class the day we covered this, and I don't see the topic discussed in Gouvêa's book, so all I have is a lone equation from a classmate's notes:
$$\#\{\text{tame Galois extensions } K/\mathbb{Q}_p \, : \, \operatorname{Gal}(K/\mathbb{Q}_p) \cong G\} = \frac{\#\{(a,b) \in G \times G \, : \, aba^{-1}=b^{p},\, \langle a,b \rangle = G\}}{\left|\operatorname{Aut}(G)\right|}$$
I know that $\left|\operatorname{Aut}(\mathcal{Q}_8)\right| = \!\left|\mathcal{S}_4\right| = 24$, so when $p\equiv 1,2 \!\pmod{4}$, I get that there are $0$ such Galois extensions, and when $p\equiv 3 \!\pmod{4}$, I get that there is only $1$. This seems odd to me.
Where does this equation come from? Is it even correct? I have tried to Google it, but have thus far been unsuccessful.