Finite extensions of groups Let $G$ be a group which is a finite extension of the group $H$.
As far as I know, by definition, this means that there is a finite normal subgroup $N$ of $G$ such that $G/N=H$. 
But, is it equivalent to say that there exists a finite index subgroup of $G$ which is isomorphic to $H$?
Moreover, I'm interested in the following: Assume that $N$ is a finite normal subgroup of $G$ and that $Z^d$ has finite index in $G/N$.  Is $G$ a finite extension of $Z^d$? 
So far, both questions have been answered negatively.
What about the following:
If $G$ is a finite extension of $H$, does there exist a finite index subgroup of $G$ which is isomorphic to $H$?
Secondly, if $N$ is a finite normal subgroup of $G$ and $Z^d$ has finite index in $G/N$, does there exist a finite index subgroup of $G$ which is isomorphic to $Z^d$?

The symmetric group $G=S_3$ of degree $3$ contains a cyclic group $C$ of order $3$, yet it is not an extension of $C$. Indeed, there is no normal subgroup in $G$ of index $3$ or, equivalently, of order $2$.
Next, let $\newcommand\ZZ{\mathbb{Z}}G=\langle x,y:y^2=yxyx=1\rangle$, the infinite dihedral group. The subgroup $\langle x\rangle$ is infinite cyclic of finite index, so we can take $N=\langle 1\rangle$, the trivial subgroup of $G$. Then $G$ is not an extension of $\mathbb Z$.
