Let $D_8 := \langle a,b \mid a^4 = 1 = b^2, bab = a^{-1}\rangle$
I'm trying to formally show that
$$D_{8} \cong C_4 \rtimes C_2 = \langle s\rangle \rtimes \langle t \rangle$$
My book gives as hint to consider the map $\phi: C_2 = \langle t \rangle \to Aut(C_4) = \{id, .^{-1}\}\cong C_2 $ that sends $1 \mapsto id, t \mapsto .^{-1}$ and then show that
$$D_8 \cong C_4 \rtimes_\phi C_2$$
The right-action associated with this map is given by:
$$*: C_4 \times C_2 \to C_4:$$
where $s*t =(s)((t)\phi))$ (I follow the convention in Isaac's algebra book where I write $(x)f$ instead of $f(x)$).
I tried to define an isomorphism explicitly:
$$D_8 \to C_4 \rtimes_\phi C_2$$
by mapping
$$a^jb^k \mapsto (k,j) \in \mathbb{Z}_2 \times \mathbb{Z}_4 \cong C_2 \times C_4$$
where $j \in \{0, \dots, 3\}, k = 0,1$.
This seems to be the only sensible choice for such a map.
How can I show that this map is a bijection (I see that injection or surjection is sufficient, and I also know that this is well-defined), and thus an isomorphism? Is there an easier way to see the isomorphism?
