[proofplan]
The three claims — homomorphism, kernel, surjectivity — each reduce to a single direct verification using the structure of the [Quotient Group](/theorems/790). The homomorphism property is read off the coset multiplication rule $(gN)(hN) = (gh)N$ that defines the operation on $G/N$. The kernel computation uses the fact that the identity of $G/N$ is the coset $N$ itself, and that $gN = N$ exactly when $g \in N$. Surjectivity is immediate from the definition of $G/N$ as the set of all cosets $\{gN : g \in G\}$.
[/proofplan]
[step:Verify that $\pi$ respects the group operation]
Declare the map under consideration as
\begin{align*}
\pi: G &\to G/N \\
g &\mapsto gN.
\end{align*}
Let $g, h \in G$. By the [Quotient Group](/theorems/790), the binary operation on $G/N$ is the well-defined coset multiplication $(gN)(hN) = (gh)N$, where well-definedness uses the hypothesis $N \trianglelefteq G$. We compute
\begin{align*}
\pi(gh) &= (gh)N = (gN)(hN) = \pi(g)\,\pi(h).
\end{align*}
Since $g, h$ were arbitrary, $\pi$ is a group homomorphism.
[/step]
[step:Identify the kernel as $N$]
The identity element of $G/N$ is the coset $eN = N$, where $e$ is the identity of $G$. By definition of the kernel,
\begin{align*}
\ker \pi &= \{g \in G : \pi(g) = N\} = \{g \in G : gN = N\}.
\end{align*}
We now show $\{g \in G : gN = N\} = N$ by double inclusion.
For the inclusion $N \subseteq \ker\pi$: if $g \in N$, then for every $n \in N$ we have $gn \in N$ (closure of $N$ under the group operation), so $gN \subseteq N$. Conversely, for $n \in N$, the element $g^{-1}n$ lies in $N$ (since $g \in N$ implies $g^{-1} \in N$), so $n = g(g^{-1}n) \in gN$, giving $N \subseteq gN$. Hence $gN = N$.
For the inclusion $\ker\pi \subseteq N$: if $gN = N$, then in particular $g = ge \in gN = N$, so $g \in N$.
Combining the two inclusions, $\ker \pi = N$.
[/step]
[step:Read off surjectivity from the definition of $G/N$]
By the [Quotient Group](/theorems/790), the underlying set of $G/N$ is
\begin{align*}
G/N &= \{gN : g \in G\}.
\end{align*}
Thus every element of $G/N$ has the form $gN = \pi(g)$ for some $g \in G$, and $\pi$ is surjective. This completes the verification of all three properties.
[/step]