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 $\operatorname{ker} f=N⊆H$

Answer 3

I have come up with a simple, non-constructive proof, inspired by this post.

Let $ X = \{ gH \ | \ g \in G \}$ be the set of left cosets of $H$ in $G$.

Define $\varphi : G \to S_X$, where $S_X$ is the symmetric group of $X$, by $$\varphi(g)(xH) = gxH$$

It is trivial to show that $\varphi$ is well defined and indeed is a homomorphism:

$$\varphi(ab)(xH) = abxH = \varphi(a)(\varphi(b)(xH)) = (\varphi(a)\circ\varphi(b))(xH) $$

Note that:

1. $\ker\varphi$ is normal in $G$. This is obvious, because kernel of any homomorphism is normal.

2. $\ker\varphi \subseteq H$. If $a \in \ker\varphi$, then $\varphi(a)=id$, so $xH=\varphi(a)(xH)=axH$ for all $x\in G$. This implies $x^{-1}ax\in H$ (cosets $aH$ and $bH$ coincide iff $a^{-1}b\in H$). This is true for $x\in H$ in particular, so $a\in H$.

3. $[G:\ker\varphi]<\infty$, because by isomorphism theorem, we obtain: $$[G:\ker\varphi] = \left| G / \ker\varphi \right| \stackrel{\text{iso}}{=} \left| \varphi(G) \right| < |S_X| < \infty.$$

isomorphism theorem visualisation:

This ends the proof, because $\ker\varphi$ is normal in $G$, subset of $H$ and has finite index in $G$.

Please note that this proof does not specify the exact value of $\ker\varphi$, although it is relatively simple to show that $\ker\varphi = \bigcap_{x\in G} xHx^{-1}$.