If $H \leq G$ has finite index, then $G$ has a normal subgroup of finite index [duplicate]

From a practice test for a Masters Qual. Exam:

Let $H$ be a subgroup of finite index in a group $G$. Prove that $G$ has a normal subgroup $K$ of finite index with $K \subseteq H$.

It makes perfect sense intuitively that $K$ should exist, I just don't know how to get ahold of it.

• Trivial example: Trivial (sub)group. :P Jan 6 '17 at 1:34
• @ajotatxe: Just realized that myself. Jan 6 '17 at 1:35
• I think what you're looking for is the meet (in the subgroup lattice) of the conjugacy class of subgroups containing $H$. That should be your $K$ (it will be normal; why?). Jan 6 '17 at 1:37
• @JustinBenfield It will be normal because it will include everything in a conjugacy class and no more--thus it is invariant under conjugation! But does that necessarily have finite index? Jan 6 '17 at 1:39
• Does this answer your question? For $G$ group and $H$ subgroup of finite index, prove that $N \subset H$ normal subgroup of $G$ of finite index exists Dec 21 '21 at 3:37

Let $x_1, ... , x_n$ be a set of distinct left coset representatives for $H$ in $G$, and let

$$K = \bigcap\limits_{i=1}^n x_iHx_i^{-1}$$

This is a subgroup of finite index in $G$, because the intersection of two subgroups of finite index is still of finite index.

To show that $K$ is normal in $G$, note that if $g \in G$, then $gx_1, ... , gx_n$ is another set of left coset representations for $H$ in $G$. Hence there is a permutation $\sigma$ such that $gx_iH = x_{\sigma(i)}H$. Then also $Hx_i^{-1}g^{-1} = Hx_{\sigma(i)}^{-1}$, and so

$$gx_iHx_i^{-1}g^{-1} = x_{\sigma(i)}Hx_i^{-1}g^{-1} = x_{\sigma(i)}Hx_{\sigma(i)}^{-1}$$

Thus $$gKg^{-1} = \bigcap\limits_{i=1}^n gx_iHx_i^{-1}g^{-1} = \bigcap\limits_{i=1}^n x_{\sigma(i)}Hx_{\sigma(i)}^{-1} = K$$

You can obtain the same subgroup $K$ less tediously by defining it to be the kernel of the homorphism $\phi: G \rightarrow \textrm{Sym}(G/H)$ of $G$ into the group of bijections of the set of left cosets of $H$ in $G$, where $\phi(g)[xH] = gxH$ for all $g \in G$ and left cosets $xH$.