We proved the following problem today.

Let $G$ be a group and let $H$ be a subgroup of finite index.

Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.

One asked if there is a non-trivial example of this problem.

Here I consider the followings are trivial examples.

1.An abelian group $G$.

2.A finite group $G$.

3.The subgroup $H$ itself is normal.

So the question is to give an example of a group $G$ that is an infinite non abelian group and $G$ has a proper non-normal subgroup $H$ of finite index.

Please give examples below.

Here is my suggestion.

Let $D_8$ be the dihedral group of order $8$.

\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\]
Define a homomorphism $\phi:D_8 \to \Aut(\R^2)$ as follows.

$\phi(r)$ rotates points of $\R^2$ $90^{\circ}$ centered at the origin and $\phi(s)$ is a reflection along $x$ axis.

We consider the semidirect product $G=R^2 \rtimes_{\phi} D_8$. This is an infinite non-abelian group.

Consider the subgroup of $H$, whose underlying set is $\R^2\times \langle s \rangle$. Then $H$ is not normal in $G$ and has index $[G:H]=4$.

Therefore, this serves as a nontrivial example.