Double cover of symmetric group

Definition

The term double cover of symmetric group is used for a stem extension where the base normal subgroup is cyclic group:Z2 and the quotient group is an symmetric group of finite degree.

A double cover exists for the symmetric group $S_{n}$ only when $n\geq 4$. Further, for each $n$, there are two possibilities for the double cover, the "+" type and the "-" type.

If we consider the group cohomology of symmetric groups and in particular the second cohomology group for trivial group action $H^{2}(S_{n};\mathbb {Z} _{2})$ for $n\geq 4$, this second cohomology group is isomorphic to the Klein four-group, and two of the three nontrivial elements of the group correspond to these double covers. The third nontrivial element corresponds to a group extension where the alternating group in the quotient is kept intact and the odd permutations square to the non-identity central element.

In all cases, both double covers are Schur covering groups for the corresponding symmetric group, because the Schur multiplier is precisely cyclic group:Z2.

Presentation for "-" type double cover

This group, denoted $2\cdot S_{n}^{-}$, has a presentation with generating set of size $n$ given by $z,s_{1},s_{2},\dots ,s_{n-1}$. The idea is that under the surjective map to $S_{n}$, $z$ maps to the identity and the relations collapse to the Coxeter presentation of $S_{n}$. The subgroup $\langle z\rangle$ is the base cyclic group:Z2. The relations (here $e$ denotes the identity element):

Relation	Condition on subscripts	Number of such relations	What it descends to when we quotient to $S_{n}$
$\!z^{2}=e$	no subscripts	1	$e=e$, i.e., a vacuous relation
$\!s_{i}^{2}=z$	$1\leq i\leq n-1$	$n-1$	$s_{i}^{2}=e$
$\!s_{i+1}s_{i}s_{i+1}=s_{i}s_{i+1}s_{i}z$	$1\leq i\leq n-2$	$n-2$	$s_{i+1}s_{i}s_{i+1}=s_{i}s_{i+1}s_{i}$
$\!s_{j}s_{i}=s_{i}s_{j}z$	$1\leq i<j\leq n-1$ and $|i-j|\geq 2$	$(n-2)(n-3)/2$	$s_{j}s_{i}=s_{i}s_{j}$
Total (--)	--	$(n^{2}-n+2)/2$	--

Presentation for "+" type double cover

This group, denoted $2\cdot S_{n}^{+}$, has a presentation with generating set of size $n$ given by $z,s_{1},s_{2},\dots ,s_{n-1}$. The idea is that under the surjective map to $S_{n}$, $z$ maps to the identity and the relations collapse to the Coxeter presentation of $S_{n}$. The subgroup $\langle z\rangle$ is the base cyclic group:Z2. The relations (here $e$ denotes the identity element):

Relation	Condition on subscripts	Number of such relations	What it descends to when we quotient to $S_{n}$
$\!z^{2}=e$	no subscripts	1	$e=e$, i.e., a vacuous relation
$\!s_{i}z=zs_{i}$	$1\leq i\leq n-1$	$n-1$	$s_{i}=s_{i}$, i.e., a vacuous relation
$\!s_{i}^{2}=e$	$1\leq i\leq n-1$	$n-1$	$s_{i}^{2}=e$
$\!s_{i+1}s_{i}s_{i+1}=s_{i}s_{i+1}s_{i}$	$1\leq i\leq n-2$	$n-2$	$s_{i+1}s_{i}s_{i+1}=s_{i}s_{i+1}s_{i}$
$\!s_{j}s_{i}=s_{i}s_{j}z$	$1\leq i<j\leq n-1$ and $|i-j|\geq 2$	$(n-2)(n-3)/2$	$s_{j}s_{i}=s_{i}s_{j}$
Total (--)	--	$(n^{2}+n)/2$	--

Particular cases

$n$	order of symmetric group $S_{n}$ equals $n!$	order of double cover of symmetric group = $2(n!)$ (twice the preceding column)	symmetric group $S_{n}$	$2\cdot S_{n}^{-}$ (double cover of "-" type)	$2\cdot S_{n}^{+}$ (double cover of "+" type)	Cohomology information	Cohomology group information
4	24	48	symmetric group:S4	binary octahedral group	general linear group:GL(2,3)	group cohomology of symmetric group:S4	second cohomology group for trivial group action of S4 on Z2
5	120	240	symmetric group:S5	double cover of symmetric group:S5 of minus type	double cover of symmetric group:S5 of plus type	group cohomology of symmetric group:S5	second cohomology group for trivial group action of S5 on Z2