0
$\begingroup$

Apologies if this is elementary, but I have never heard the terminology before:

What is a "non-splitting covering" of a finite group?

I encountered the term while reading this paper, in which the author introduces a group $H$ with subgroups $V$ and $T$ with the presentation $$H = \langle a,b,x,y \mid a^4=b^4=x^4=y^3=1, a^2=b^2=x^2, [b,a]=a^2, [x,a]=[y,b]=ab,$$ $$ [y,a]=b^3,[x,y]=y^2,[x,b]=a^3b\rangle;$$ $$V = \langle a,b \rangle = \mbox{Fit}H\text{ and }T = \langle x, y \rangle.$$ She then concludes that $H = VT$ is the "non-splitting covering of $S_4$", with $V \cong Q_8$ and $H/V \cong S_3$. I am not sure what she means by this; my best guess is that $H$ contains $S_4$ and is not a split extension. As a side note, if anyone knows where to find a more intuitive description of (or a more common name for) this group, I would be appreciative.

$\endgroup$
1
  • 2
    $\begingroup$ The term I have usually seen is 'non-split cover'. A cover of a group $G$ is just a group that has a quotient isomorphic to $G$. $\endgroup$
    – Colin Reid
    Commented Aug 23, 2012 at 1:06

1 Answer 1

3
$\begingroup$

I think it is rather that $H$ has a homomorphic image $S_4,$ but no subgroup $S_{4}.$ It is sometimes called a double cover of $S_{4}.$ The group $S_{4}$ has two such non-isomorphic (proper) double covers. One is ${\rm GL}(2,3)$ which has a semidihedral Sylow $2$-subgroup. The other is the binary octahedral group, which has a generalized quaternion Sylow $2$-subgroup. The latter group is a subgroup of ${\rm GL}(2,9)$ which may be obtained from ${\rm GL}(2,3)$ by replacing elements by scalar multiples of determinant $1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .