Group names in GAP character table library

abstract-algebracomputer-algebra-systemsfinite-groupsgapgroup-theory

Let $G = M_{12}.2$, automorphism group of the Mathieu group $M_{12}$.

In GAP, with DisplayAtlasInfo("M12.2");; we get information about G as listed here or here.

One of the maximal subgroups is listed as having structure D8.(S4x2).

So in ATLAS notation, I think this would mean that the subgroup has a normal subgroup $D_8$ with quotient $S_4 \times C_2$.

But as far as I can tell by computing with GAP, this maximal subgroup does not have a dihedral group $D_8$ as a normal subgroup. It does have $C_2^3$ as a normal subgroup, with quotient $S_4 \times C_2$. Also it seems in the ATLAS this maximal is denoted by $(2_{+}^{1+4}:S_3).2$, for example here.

Is this correct and what does D8.(S4x2) mean in GAP? More generally how to interpret the group names used in GAP?

Best Answer

This is a misprint in the library (that then propagated to web pages you refer to that just pull data from the library).

In the printed version of the ATLAS the group is an extension of one named $M_8.S_4$, here $M_8$ is a Mathieu-type name for $Q_8$. However the extra 2 does not preserve the $Q_8$, which makes the name $M_8.S_4.2$ misleading. Better use the name from the ATLAS webpages you cite. An alternative name could be a non-split extension $2^3\cdot (2\times S_4)$.

This will be corrected in a future release.

Related Solutions

Understanding Direct product of subgroups in GAP

The group $\operatorname{GL}_2(3)$ has no subgroup isomorphic to $C_2 \times D_8$. If you take the normal $C_2$ copy, you won't find any $D_8$ copies intersecting it trivially. E.g.

gap> gl:=GL(2,3);
GL(2,3)
gap> List(List(ConjugacyClassesSubgroups(gl),Representative),x->StructureDescription(x));
[ "1", "C2", "C2", "C3", "C4", "C2 x C2", "S3", "S3", "C6", "Q8", "D8", "C8", "D12", "QD16", "SL(2,3)", "GL(2,3)" ]

Compute character table of $S_4$ in Fulton-Harris

The observation is that, since $(12)(34)$ maps to the identity, then it's in the kernel of $S_4\to \operatorname{Aut}(W)$, so all three conjugates of $(12)(34)$ must be in the kernel as well (in addition to $1$ of course).

This set $V$ is a union of conjugacy classes, so it's a normal subgroup. Whenever you have a normal subgroup in the kernel of a homomorphism, that homomorphism factors through the quotient: $$S_4 \twoheadrightarrow S_4/V \to \operatorname{Aut}(W)$$ (The universal factoring is through the quotient by the kernel itself.) What this means is that this $W$ representation may be regarded as coming from $W$ as a representation of the smaller group $S_4/V$.

The question now is which group is $S_4/V$? Let $C$ be the conjugacy class of $(12)(34)$, and consider the action of $S_4$ on $C$ by conjugation. A group action, recall, is a homomorphism $S_4\to \operatorname{Bij}(C)$, and since $C$ has three elements, the group of bijections $\operatorname{Bij}(C)$ is isomorphic to $S_3$ by choosing some bijection between $C$ and $\{1,2,3\}$. Thus, this group action is "actually" a homomorphism $S_4\to S_3$. What is its kernel? You can check that $V$ is an abelian subgroup, so it acts on each element of $C$ trivially, hence $V$ is in the kernel. You can also check that all six bijections of $C$ can be realized by conjugating by some element of $S_4$, so $S_4\to\operatorname{Bij}(C)$ is surjective. Since $\lvert V\rvert =4$, $\lvert S_4\rvert=4!$, and $\lvert S_3\rvert=3!$, it follows that $K$ is the kernel, so by the first isomorphism theorem $S_4/V\cong S_3$.

One way I like to think about this is to imagine a tetrahedron with vertices labeled 1-4, and to think about the elements of $C$ as being pairs of opposite edges, so for example $(12)(34)$ is the pair of edges connecting vertices 1&2 and vertices 3&4. Conjugations by $S_4$ correspond to rotating and reflecting this tetrahedron (a conjugation relabels the vertices arbitrarily: rotate and reflect to bring the tetrahedron back to "standard" position). Then you can realize that the three sets of opposite edge pairs have the property that they meet at each vertex. The data of the $S_4$ action on $C$ reduces to how the three edges at vertex 1 are permuted.

Best Answer

Related Solutions

Understanding Direct product of subgroups in GAP

Compute character table of $S_4$ in Fulton-Harris

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