Number of distinct conjugates of a subgroup is finite Let $H$ be a subgroup of $G$.  I would like to prove that if $H$ has finite index in $G$, then there is only a finite number of distinct subgroups in $G$ of the form $aHa^{-1}$.  (This is an exercise in [Herstein, Topics in Algebra], in the section on subgroups and Lagrange's theorem.)
The assertion clearly holds if $|H|$ (or $|G|$) is finite.  Also, the number of distinct conjugates $aHa^{-1}$ is 1 if $H \trianglelefteq G$.  So we need to consider only the case $|H|=\infty, H \ntrianglelefteq G$. I am also wondering what would be some specific examples of such infinite groups $H \le G$ with $|G:H| < \infty$ and $H \ntrianglelefteq G$.
 A: Are you familiar with the concept of group actions? The group $G$ acts on the conjugacy class of $H$ via the conjugation action. There is only one orbit, and its stabilier is $N_G(H)$, hence the orbit is in (equivariant) bijection with the coset space $G/N_G(H)$ via $gN_G(H)\longleftrightarrow gHg^{-1}$.
But $H\le N_G(H)\le G$ so $N_G(H)$ has finite index, so $G/N_G(H)$ is finite.
The class of profinite groups exhibits a large number of examples of groups with subgroups of finite index - in fact they have topology which is defined in terms of them.
A: I want to give an explicit answer to the last part of your question, as otherwise this is pretty much a duplicate...
So, what is a group which has a non-normal finite-index subgroup? Take an arbitrary infinite group $H$, for example the infinite cyclic group $H=C_{\infty}\cong \mathbb{Z}$, and take its cross-product with a non-abelian finite simple group $S$, for example $S=A_5$. Form the group $G=H\times S$.
As $S$ is non-abelian it has a proper, non-trivial subgroup $K$. Then the subgroup $H\times K\leq G$ is not normal in $G$, but does have finite index.
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 19 on p.48.
I solved this problem as follows:

$$aHa^{-1}=bHb^{-1}\Leftrightarrow (b^{-1}a)H(b^{-1}a)^{-1}=H\\\Leftrightarrow b^{-1}a\in N(H):=\{a\in G\mid aHa^{-1}=H\}.$$
See Problem 16 on p.47 about the subgroup $N(H)$ of $G$.
Obviously, $N(H)\supset H$.
By assumption, $i_G(H)=\#G/H<+\infty$.
So, $i_G(N(H))<+\infty$.
Note that $i_G(N(H))$ is equal to the number of distinct subgroups in $G$ of the form $aHa^{-1}$.

