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What are the necessary and sufficient conditions for a group $H$ to be a central extension of the quaternions $Q_8$ ?

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You need to define what you mean by one group being an extension of another. I can think of at least three possible meanings. – Derek Holt Jan 2 '13 at 9:37
Sally, it is in general not useful to delete comments. (For example, while still understandable, the first sentence of the answer now, since your clarifiying comment is gone, seems slightly strange.) The idea to avoid clutter (which is I assume behind it) is in some sense good, however it can have negative side-effect; thus I would like to ask you to rather leave comments around. – quid Jan 2 '13 at 14:55
@Quid: I have edited the question and replaced "extension" by "central extension". I was planning to delete the first sentence of my answer, but that would have rendered your comment confusing, so I didn't! – Derek Holt Jan 2 '13 at 15:33

1 Answer 1

up vote 6 down vote accepted

So you mean "central extension". Let $H/K \cong Q_8$ with $K \le Z(H)$. Since $Q_8$ has trivial Schur Multiplier, $K \cap H' = 1$, so $|H'| = 2$, and $H$ is a subdirect product of an abelian group $L$ with $Q_8$, where $L$ and $Q_8$ have the common quotient $V = C_2^2$ (the Klein 4-group).

More precisely, let $L$ be an abelian group with an epimorphism $\phi:L \to V$ and let $\psi:Q_8 \to V$ be an epimorphism. Then $H = \{ (g,h) \mid g \in L, h \in Q_8, \phi(g)=\psi(h) \}$.

Further explanation follows. First note that $Q_8$ has a presentation with two generators and two relations, namely $\langle x,y \mid x^2=y^2, y^{-1}xy=x^3 \rangle$.

Let $G$ be a any finite group with a presentation with $n$ generators and $n$ relations for some $n$. So we can write $G=F/R$ with $F$ free of rank $n$, and $R$ the normal closure in $F$ of $n$ elements of $F$. So $R/[R,F]$ is an $n$-generated abelian group. Now $R[F,F]/[F,F]$ has finite index in the rank $n$ free abelian group $F/[F,F]$, and so it is also free abelian of rank $n$. But $R[F,F]/[F,F] \cong R/(R \cap [F,F])$, so the $n$-generated abelian group $R/[R,F]$ has the rank $n$ free abelian quotient $R/(R \cap [F,F])$. Hence $(R \cap [R,F])/[R,F]$ (which is the Schur Multiplier of $G$) is trivial.

Now we can lift the epimorphism $F \to Q_8$ to a homomorphism $\rho:F \to H$. Since $K \le Z(H)$ and $H = K {\rm Im}(\phi)$, this implies that $H' \le \phi([F,F])$, and so $H' \cap K \le \phi(R \cap [F,F])$. Note that $K \le Z(H)$ implies that $[R,F] \le {\rm Ker}(\phi)$, and so $[R,F] = R \cap [F,F]$ implies $H' \cap K = 1$.

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That's easy to prove more directly. Since the inverse images in $H$ of the cyclic subgroups of order 4 in $Q_8$ are all abelian, the centralizer of the inverse image of the eleemnt of order 2 in $Q_8$ must be the whole of $H$. – Derek Holt Jan 2 '13 at 14:00
Sally, by deleting your comment, you have made my comment incomprehensible! My comment is about showing that there is no group $H$ with $H/Z(H) \cong Q_8$. – Derek Holt Jan 2 '13 at 15:28
@Derek: Agreed. sorry. I thought the question is trivial. That is why I deleted my comment. But that leads me to another (non-trivial) question: Do you know any necessary and sufficient condition on a group $G$ to has the form $\frac{H}{Z(H)}$ ? – user30300 Jan 3 '13 at 12:31
@Sally: You can find some information on such groups here:… – Ralph Jan 3 '13 at 19:34

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