For $G$ group and $H$ subgroup of finite index, prove that $N \subset H$ normal subgroup of $G$ of finite index exists

Question

Let $G$ be a group and $H$ be a subgroup of $G$ with finite index. I want to show that there exists a normal subgroup $N$ of $G$ with finite index and $N \subset H$. The hint for this exercise is to find a homomorphism $G \to S_n$ for $n := [G:H]$ with kernel contained in $H$.

The standard solution suggests to choose $\varphi$ as the homomorphism induced by left-multiplication $\varphi: G \to S(G/H) \cong S_n$. I'm not 100% sure if I understand this correctly. What exactly does $\varphi$ do? We take $g \in G$ and send it to a bijection $\varphi_g: G/H \to G/H, xH \mapsto gxH$? If so, how can I see that its kernel is contained in $H$? Also, the standard solution claims its image is isomorphic to $G/N$ and thus $N$ has a finite index in $G$, how can I see that the image is isomorphic to $G/N$?

Thanks in advance for any help.

Answer 1

Your definition of $\varphi$ looks fine. Anything in the kernel must in particular fix $H$, and $gH = H$ is equivalent to $g \in H$. On the other hand I think $N = \ker \varphi$ can be a proper subgroup of $H$. As an example, which is silly because the group is finite, if you take $G = S_3$ and $H = \{1, (12)\}$ then this process produces $N = \{1\}$.

For the second question, this is just the "first" isomorphism theorem.

Answer 2

I don't know it's true or false but I try this as this

$H$ is a subgroup of $G$. $(G:H)=n$, we can get atleast one normal subgroup $N⊆H$.

Let $［G:H］=\{g_1,g_2,...,g_n\}$

Now we define a mapping $f:G \to S_n$ such that $f(a)=g_i$ where $a∈g_i, N⊆g_iH$ Clearly mapping is well defined.

Let $f(b)=g_j$ where $g_j∈g_j, N ⊆g_jH$.

Now $a∈g_i N$ and $b ∈g_j N$ therefore $ab∈g_i.g_jN⊆g_ig_jH$ Therefore $f(ab)=gi.gj=f(a)f(b)$, $f$ is homomorphism.

Let $x∈\operatorname{ker}f$ .Then $x∈N⊆H$ i.e $\operatorname{ker} f=N⊆H$