Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
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 '13 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 '13 at 14:25
how do you show it is homomorphism? – user2993422 Sep 14 '15 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 – user2993422 Sep 14 '15 at 9:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.