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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.

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    Trivial example: Trivial (sub)group. :P – 2017-01-06
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    @ajotatxe: Just realized that myself. – 2017-01-06
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    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?). – 2017-01-06
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    @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? – 2017-01-06
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    One way to go about it is to let $G$ act on $G/H$ via left multiplication. Then the kernel of this action $\phi\colon G\to S(G/H)$ is normal and contained in $H$. It'll have finite index since the quotient is isomorphic to a subgroup of $S(G/H)$. – 2017-01-06
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    I meant meet, not join. – 2017-01-06
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    @JustinBenfield. Ok, I'm not sure why that would have finite index or why it would be normal though. – 2017-01-06
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    idk about finite index myself, but normality follows from the fact that the meet of the subgroups in the conjugacy class containing $H$ is precisely the intersection of all of those groups, and that set is invariant (as a set) under conjugation because it is contained in every conjugate of $H$. – 2017-01-06
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    @BenWest what do you mean by $S(G/H)$? – 2017-01-06
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    @setholopolus Pretty sure he means the symmetric group on the set of the group $G/H$. – 2017-01-06
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    ^What Justin said, just the group of set bijections on the finite set $G/H$. – 2017-01-06
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    @BenWest: What if $H$ isn't normal? Then you need to replace $G/H$ with the set of *left* cosets? Right? – 2017-01-06
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    @JustinBenfield It's okay if $H$ isn't normal. Although $G/H$ is not a group, it still forms a set of (left) cosets, and left translation will still induce a bijection of this set. There is no need for $G/H$ to carry a group structure for $G$ to act on it in this way. – 2017-01-06
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    Oh, I was implicitly assuming $G/H$ is the set of left cosets, not right cosets, thanks. – 2017-01-06
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    @BenWest Thanks! I think I get it. If you write it as an answer with the clarifications included, I'll be sure to accept it. – 2017-01-06

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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 $xH$.