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

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.

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.

• For the third statement of the first isomorphism theorem, am I correct that this just follows from the universal property of factor groups?
– Huy
Jun 24, 2013 at 14:09
• @Huy I certainly think it's part of that circle of ideas. For me the universal property of $G/H$, where $H$ is a normal subgroup of $G$, is that any homomorphism $f\colon G \to G'$ with $H \subset \ker f$ factors uniquely through it. It seems like it's an extra step to say that if $H = \ker f$ then you get an embedding.
– TTS
Jun 24, 2013 at 14:25
• how do you show it is homomorphism? Sep 14, 2015 at 9:52
• and also i don't understand why N is in H? I think that from your solution we can assume N in xH Sep 14, 2015 at 9:59
• How can we show $S(G/H) \cong S_n$? Jan 17, 2018 at 9:30

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 $\operatorname{ker} f=N⊆H$

• This is a question asked four+ years ago. "I don't know if it's true or false but I try this....". Don't post "guesses, conjectures, or anything that could be misleading to an asker. Nov 3, 2017 at 17:29