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 $Z^d$?
 A: 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 $\mathbb Z$.
A: Unfortunately, the concept "extension of $N$ by $H$" for groups $N$ and $H$ is used by some authors to mean a group with normal subgroup $N$ with quotient $H$ and by others (the split is roughly 50/50 as far as I can judge) to mean group with normal subgroup $H$ and quotient $N$.
So "finite extension of $H$" could be interpreted in either of the two ways you suggested. In a given context, you can usually work out what the author means by reading some of the text.
The answer to your second question: "If N is a finite normal subgroup of G and ${\mathbb Z}^d$ has finite index in G/N, does there exist a finite index subgroup of G which is isomorphic to ${\mathbb Z}^d?$" is yes.
To prove this, we can assume that $N \lhd G$ with $G/N \cong {\mathbb Z}^d$. Since $C_G(N)$ has finite index in $G$, we can assume that $C_G(N)=G$, so $N$ is central in $G$. Let $N$ have exponent $e$. Then, if $x_1,\ldots,x_d$ are such that $x_iN$ freely generate the abelian group $G/N$, the commutators $[x_i^e,x_j^e]$ are all trivial, so the subgroup $\langle x_1^e,\ldots,x_d^e \rangle$ has finite index in $G$ and is isomorphic to ${\mathbb Z}^d$. 
