Let $G$ be a group, and $H$ a subgroup of G, and f in $Aut(G)$. Prove that $[G:H]=[G:f(H)]$.
If $N$ is a normal subgroup of $G$ and $f$ in $Aut(G)$, prove that $G/N$ is isomorphic to $G/f(N)$.
My ideas so far are the following:
For 1. consider $f|_H: H \rightarrow f(H)$. This is a bijection, so $\# N = \# f(N)$. Hence by Lagrange's Theorem $[G:H] = \frac{\#G}{\#N} = \frac{\#G}{\#f(N)} = [G:f(H)]$.
For 2. Define the homomorphism $h: G \rightarrow G/f(N) : g \mapsto gf(N)$. Then $a \in Ker(f) \Leftrightarrow h(a) = f(N) \Leftrightarrow af(N) = f(N) \Leftrightarrow a \in N$. Then by the isomoprhism theorem we have that $G/N \cong G/f(N)$.
Does anyone know in how far it is correct what I have now? The troubling thing for 2. is that I'm not using the assumption that $N$ is a normal subgroup.
Call $f(N)=K$. Since f is an automorphism it maps a normal subgroup to a normal subgroup. Hence K is normal. Actually any surjective group homomorphism maps a normal subgroup to a normal subgroup.
Define the map $h:G \to G/K$ as h(g)=f(g)K
Observe that the map is well defined. It is a group homomorphism. And now Check its kernel.You will get $Ker(h)=N$, then by first isomorphism theorem $G/N \cong G/K$