most recent 30 from http://mathoverflow.net 2013-06-20T00:24:49Z http://mathoverflow.net/feeds/question/105255 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105255/what-is-a-non-splitting-covering-of-a-finite-group Alexander Gruber 2012-08-22T17:45:46Z 2012-08-22T21:46:09Z

<p>Apologies if this is elementary, but I have never heard the terminology before: </p> <blockquote> <p>What is a <strong>"non-splitting covering"</strong> of a finite group?</p> </blockquote> <p>I encountered the term while reading <a href="http://www.bdim.eu/item?fmt=pdf&amp;id=BUMI_2002_8_5B_1_131_0" rel="nofollow">this paper</a>, 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.</p>

http://mathoverflow.net/questions/105255/what-is-a-non-splitting-covering-of-a-finite-group/105271#105271 Geoff Robinson 2012-08-22T21:46:09Z 2012-08-22T21:46:09Z

<p>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 $1$.</p>