Subgroup automorphisms

Question

$N$ is a subgroup of $G$, $f \in \operatorname{Aut}(G)$ and $f(N)=N$. Define $h(gN)=f(g)N$, $g \in G$. Is $h\in \operatorname{Aut}(G/N)$ ?

The problem is when I'm trying to prove formally that $h$ is an injection, I come to the part when I need to show that $f(g_1)N=f(g_2)N$ implies $g_1N=g_2N$. Can I conclude that, because $f$ is an injection, it follows immediately .. ?

Answer 1

Suppose $gN \in \ker h$. Then, $f(g) \in N$. Since $f(N)=N$, $f$ is an automorphism when restricted to $N$. It follows that $g \in N$ and so your map is injective.

Answer 2

Since $f(g_1)N = f(g_2)N$ this implies that $f(g_2) \in f(g_1)N$ v.v. by cosests. However, what does it mean for $f(g_2) \in f(g_1)$? It means there exists an $n \in N$ so that $f(g_1)n = f(g_2)$. However, $f(N) = N$ and f is an automorphism. So there exists an $n' \in N$ so that $f(g_1 n') = f(g_1)f(n') = f(g_1)n = f(g_2)$. Since f is injective, we have $g_1n' = g_2$. So $g_2 \in g_1N$. Since cosets are equivalence classes, $g_2N = g_1N$.