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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 $Z^d$?

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2 Answers 2

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 $\mathbb Z$.

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OK. I see. 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? –  Johnny Apr 8 '12 at 6:57
    
Please ask the questions in the main question. Comments are not a good place to do this. –  Mariano Suárez-Alvarez Apr 8 '12 at 7:01
    
In any case, my answer's conclusion is that you should look at simple examples! –  Mariano Suárez-Alvarez Apr 8 '12 at 7:01

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 ${\mathbb Z}^d$.

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Derek Holt: Could you explain why your assumptions are without loss of generality? 1) G/N=Z^d 2) N is central in G Thank you. –  Johnny May 19 '12 at 7:12

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