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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.

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1 Answer 1

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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.

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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

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