If $H$ is a subgroup of $G$ having finite index, then $H$ contains a normal subgroup of $G$ of finite index I'm trying to show that any subgroup $H$ of a group $G$ having finite index must contain a normal subgroup of $G$ of finite index.
I tried to define a homomorphism $\psi:G/H \to G/H$ given by $\psi(xH) =gxH$ and prove that $\ker(\psi)\subset H$. Is there another way to solve this problem?
 A: Given a subgroup $H\leq G$ of finite index, $G$ acts transitively on the set of left cosets $G/H$ via $g\cdot(g'H) := (gg')H$, for $g \in G$ and $g'H \in G/H$; the kernel $K$ of the corresponding group homomorphism $G \to \text{Sym}(G/H)$ is a normal subgroup of $G$ contained in $H$, and hence has finite index.
Comment: One can show that $K = \bigcap_{g \in G} gHg^{-1}$. This is core of $H$ in $G$; it is the largest normal subgroup of $G$ contained in $H$. 
A: Hint: Consider the induced homomorphism from $G$ to the group of permutations of $G/H$.
A: As another answer indicates, you're "close".
The problem is, the map $\psi_g: G/H \to G/H$ given by $\psi_g(xH) = (gx)H$ might not be a homomorphism, because $G/H$ might not be a group.
So is there a way to fix this?
It turns out there is, and it's common to different branches of algebra.
The map $\psi_g$ isn't necessarily a homomorphism, but it is bijective on $G/H$, for if:
$\psi_g(xH) = \psi_g(yH)$, then $(gx)H = (gy)H$, that is: $(gy)^{-1}(gx) \in H$.
Expanding this, we have: $y^{-1}g^{-1}gx = y^{-1}x \in H$, so that $xH = yH$.
This shows that $\psi_g$ is injective, and since $G/H$ is finite, it is thus also bijective. And we know the bijections on any set (even $G/H$) form a group.
So now we have two groups, $G$ and $\text{Sym}(G/H)$. So we can ask: can we find a homomorphism now?
Your task, now, is to convince yourself (this means prove) that $g \mapsto \psi_g$ is the homomorphism we're after. Let's call this homomorphism $\phi$ (so that $\phi(g) = \psi_g$).
Now, to finish, consider the two main objects of interest with our homomorphism $\phi$-namely, its kernel and image. Fill in the blanks:
$\text{im }\phi$ is a subgroup of a ______ group, so it is _____ .
Therefore, the kernel has ______  ______ (via the Fundamental Isomorphism Theorem).
One last question for you: have we found a suitable normal subgroup?
