Let $G=D_8\times\mathbb{Z_6}$ and $N=D_8\times\{0\}$. Prove that $G/N$ is isomorphic to $\mathbb{Z}_6$.
My attempt:
I proved that $N$ is a normal subgroup of $G$. Then we can define
$$G/N = \{gN \mid g\in G\}.$$
Let $x\in G/N$ then $x=g\cdot n$ where $g\in G$ and $n\in N$. Moreover: $g\in G$, then $g=(a,b)$ where $a\in D_8$ and $b\in \mathbb{Z}_6$.
If $n \in N$ then $n=(c,0)$ where $c$ in $D_8$ and $0\in \{0\}$.
This implies: $x=(a\cdot c, b \oplus0) = (a\cdot c,b)$
Now, we can define the function $f: G/N \rightarrow \mathbb{Z}_6$ such that $f((a\cdot c,b))=b$.
This function is clearly bijective and
$$f((a\cdot c,b))f((a'\cdot c',b'))=bb' = f((a\cdot c,bb'))$$
Is my proof correct?
Define a map
$$\begin{align} f: G& \to \Bbb Z_6, \\ (a,b) &\mapsto b. \end{align}$$
Clearly $f$ is a homomorphism from $G$ to $\Bbb Z_6$.
Observe that $N$ is the kernel of this homomorphism.
By the first isomorphism theorem, $G/N$ is isomorphic to $\Bbb Z_6$.