The following is just a follow up to my previous question. I have a finite group $H$ with 14 ordinary characters. The Schur multiplier $M(H)\cong 2^2$. Hence the group $H$ will have 3 sets of projective characters with non-trivial factor sets $\alpha^{-1}_i$ of order 2, $i=1,2,3$. How to I prove that the cardinality of each of the three sets of projective characters of $H$ with factor sets $\alpha^{-1}_i$ is strictly less than |Irr(H)|=14 . Perhaps Geoff Robinson can be of help here again. Will it suffice to say because the factor sets $\alpha^{-1}_i$ are non-trivial then it follows that the number of $\alpha^{-1}_i$ - regular classes corresponding to a set of projective characters will be less than the 14 conjugacy classes of $H$, hence the result follows? – A.L. Prins
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$\begingroup$ The group $H = (4{\cdot}S_4){:}2$ of order 192 is a maximal subgroup of $U_3(3){:}2$. The central extension $2^2{\cdot} H$ has 36 irreducible characters where 22 of these are the 3 set of projective characters with associated factors sets which are lifted to the central extension. The question is if the number of any of these sets of projective characters is always less than 14? If so, how to we go about to show this? $\endgroup$– A.L. PrinsCommented May 4, 2014 at 16:13
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$\begingroup$ I said in my last answer: let the obvious Klein $4$-group in centre of $G = 2^{2}.H$ be denoted by $V.$ Choose a non-trivial linear character $\lambda$ of $V.$ The number of irreducible characters of $G$ which cover $\lambda$ is at most $14,$ and is strictly less than $14$ if and only if there is an element $v \in V \backslash {\rm ker} \lambda$ which is a commutator in $G.$ This is routine to check with a computer algebra system, particularly if you have the character table of $G,$ as previously explained. $\endgroup$– Geoff RobinsonCommented May 4, 2014 at 17:10
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$\begingroup$ The other questions are math.stackexchange.com/questions/779460/… and mathoverflow.net/questions/163208/… $\endgroup$– Jack SchmidtCommented May 4, 2014 at 20:21
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$\begingroup$ God bless @Jack Schmidt. Send me the Gap calculations. A.L. Prins $\endgroup$– user50002Commented May 5, 2014 at 10:04
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1 Answer
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We verify that $U_3(3):2$ denotes a unique group, $\operatorname{Aut}(PSU(3,3))$. We check its maximal subgroups and verify that there is a unique conjugacy class with structure $(4\cdot S_4):2$. It has order 192 and SmallGroup identifier [192, 988].
Now we compute a SchurCover and its irreducible characters, and then check their restrictions to the Schur multiplier, as explained in my slower version of Geoff's answer.
gap> h := SmallGroup(192, 988);;
gap> f := EpimorphismSchurCover(h);;
gap> f := InverseGeneralMapping(IsomorphismPcGroup(Source(f)))*f;;
gap> z := Kernel(f);; # the Schur multiplier
gap> x := Source(f);; # the Schur cover
gap> List( Irr(z), lambda -> Number( Irr( x ), chi -> not IsZero(
> ScalarProduct( RestrictedClassFunction( chi, z ), lambda ) ) ) );
[ 17, 14 ]
gap> NrConjugacyClasses( h );
17
Here are the calculations to verify the structure of $H$:
gap> u332 := Image( IsomorphismPermGroup( AutomorphismGroup( PSU(3,3) ) ) );;
gap> Size(u332)/Size(PSU(3,3)) = 2;
true
gap> ms := Filtered( MaximalSubgroupClassReps( u332 ), h -> Size(h) = 4*24*2 );;
gap> List( ms, IdGroup ); # oops two of them!
[ [ 192, 956 ], [ 192, 988 ] ]
gap> IdGroup( SymmetricGroup( 4 ) );
[ 24, 12 ]
gap> ms := Filtered( ms, h -> ForAny( NormalSubgroups(h), n -> Index(h,n)=2 and
> ForAny( NormalSubgroups(n), k -> Size(k)=4 and IsCyclic(k) and
> IdGroup(n/k)=[24,12] ) ) );;
gap> List( ms, IdGroup ); # yay only one
[ [ 192, 988 ] ]
My earlier comment had actually identified the wrong $H$, since the other $H$ contains $S_4$ as a subgroup rather than as a subquotient. You can see its decomposition here:
gap> h := SmallGroup(192, 956);;
gap> f := EpimorphismSchurCover(h);;
gap> f := InverseGeneralMapping(IsomorphismPcGroup(Source(f)))*f;;
gap> z := Kernel(f);; # the Schur multiplier
gap> x := Source(f);; # the Schur cover
gap> List( Irr(z), lambda -> Number( Irr( x ), chi -> not IsZero(
> ScalarProduct( RestrictedClassFunction( chi, z ), lambda ) ) ) );
[ 14, 7, 8, 7 ]
gap> NrConjugacyClasses( h );
14
Here is the character table of the second $H$ with the projective characters (as in Haggarty–Humphreys 1978) at the bottom. I believe most of the projective characters are only defined universally up to sign (so you might get different signs if you recalculate everything, using a different Schur cover), but these signs are calculated consistently with a “special cocycle” so the inner product and conjugacy results of H–H (1978) apply.
answered May 5, 2014 at 16:18
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$\begingroup$ GAP has functions to efficiently compute the character table of the Schur cover of $H$ without computing the Schur cover of $H$ when the Schur multiplier is $2^2$. However, this $H$ was small enough, I just computed a Schur cover through GAP's implementation of Hopf's formula. $\endgroup$ Commented May 5, 2014 at 16:21
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$\begingroup$ Also, I think you might have the “wrong” $H$ as well. The $H$ with the structure $(4\cdot S_4):2$ has 17 classes, and its covers have 31 classes. The “other” $H$ has 14 classes, and its covers have 36 classes. The “other” $H$ is the unique one with a subgroup $S_4$, but the “right” $H$ should only have it as a subquotient. $\endgroup$ Commented May 5, 2014 at 16:27
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$\begingroup$ (But it's your group, so whichever one you want is the right one. :-) $\endgroup$ Commented May 5, 2014 at 16:27
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$\begingroup$ Thank you very much and may God bless Geoff and Jack richly for helping me to solve my problems. Jack is it possible to compute the projective characters of $4^2{:}D_{12}$ explicitly? If so can you help me in this regard. You were correct that I identify my group"s structre wrongly. The structure is indeed $4^2{:}D_{12}$. $\endgroup$– user50002Commented May 7, 2014 at 12:39
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$\begingroup$ Done. My code to do this is very primitive. It is easy to produce the characters of the Schur cover, but “restricting” them to $4^2:D_{12}$ is a little trickier for people who've never done it before today. :-) $\endgroup$ Commented May 8, 2014 at 16:19