Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This older stackoverflow question may be helpful in answering the question that I ask below, although I could not work it out.

For $n\geq 1$, let $X=\lbrace 1,2, \ldots ,n \rbrace$, $Y=X \cup (-X)$ (so that in $Y$ we have all integers between $-n$ and $n$, except zero). Let ${\mathfrak S}_Y$ denote the group of permutations on $Y$, and define a subgroup $G$ of ${\mathfrak S}_Y$ by

$$ G=\lbrace \sigma \in {\mathfrak S}_Y | \forall x\in X, \sigma(-x)=-\sigma(x) \rbrace $$

(so $G$ is the centralizer of the permutation that multiplies by $-1$). Does $G$ have a name in the literature ? (UPDATE : $G$ is the group of signed permutations on $X$, also known as the Coxeter group of type $B_n$).

We can easily decompose $G$ as a semi-direct product $ {\lbrace \pm 1 \rbrace}^n \rtimes {\mathfrak S}_X$, for on one hand there are two injective homomorphisms $i \ : \ {\lbrace \pm 1 \rbrace}^n \to G$ and $j: {\mathfrak S}_X \to G$, defined by $$ i(\varepsilon_1,\varepsilon_2, \ldots ,\varepsilon_n)(x)=\varepsilon_x x \ ({\rm for}\ \varepsilon_k=\pm 1,\ x \in X) $$ and $$ j(\sigma)(x)=\sigma(x), \ j(\sigma)(-x)=-\sigma(x) \ ({\rm for}\ \sigma \in {\mathfrak S}_X,\ x\in X) $$ and on the other hand there is a surjective homomorphism $ s \ : \ G \to {\mathfrak S}_X$, defined by $$ s(\sigma)(x)=|\sigma(x)| \ ( {\rm for}\ \sigma \in G, x\in X). $$ Since ${\sf Ker}(s)={\sf Im}(i)$ and $G={\sf Im}(i){\sf Im}(j)$, this yields a split exact sequence and hence $|G|=| {\lbrace \pm 1 \rbrace}^n | |{\mathfrak S}_X|=2^n n!$.

Then $s^{-1}({\cal A}_X)$ is a subgroup of index $2$ of $G$ because ${\cal A}_X$ is a subgroup of index $2$ in ${\mathfrak S}_X$. Are there any other subgroups of index $2$ in $G$ ?

UPDATE AT 17:30 : For $n\geq 3$, there are at least three distinct subgroups of index 2 ; they can be viewed as the kernels of three different homomorphisms $G \to \lbrace \pm 1\rbrace$.

The first homorphism is $t=\varepsilon \circ s$ (where $\varepsilon$ is the signature of a permutation of $X$). The kernel of $t$ is the subgroup already mentioned above.

The second homomorphism is $t'$, defined by $t'(\sigma)=|\sigma(X) \setminus X| {\sf mod} 2$ (this is the homomorphism suggested by "jug" in his answer below).

The third homomorphism is $t''=tt'$.

I checked with a computer that these are the only subgroups of index $2$ when $n=3,4$ or $5$.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

In this question your group is called the group of signed permutation and it is the Coxeter group $B_n = C_n$ (it shows up for example as the Weyl group of the symplectic group $\mathop{S}_{2n}(\mathbb{F_q})$).

Hint: For another subgroup of index 2 take a look at the cardinality of $\sigma(X)\cap (-X)$ for $\sigma \in G$.


EDIT: There are not more index-2-subgroups than the three you found, as the commutator subgroup $G'$ has index 4 in $G$.

An element $\sigma \in \{\pm1\}^n$ for which $\sigma(X)\cap (-X) := X_\sigma$ has even order is contained in $G'$, since you can write $X_\sigma = X_0\stackrel{.}{\cup}X_1$ as disjoint union of two sets of the same cardinality. Take any permutation $\tau \in S_n$ that exchanges $X_0$ and $X_1$, and the element $\rho\in \{\pm1\}^n$ that equals $-1$ on $X_0$ and $1$ otherwise. Then $\sigma = [\tau, \rho]$ shows that $G'\cap\{\pm1\}^n$ has index $2$ in $\{\pm1\}^n$. As the alternating group is the commutator subgroup of the symmetric group (of index $2$), $G'$ has index at most $4$ in $G$, and the homomorphisms you found show that the index is exactly $4$.

share|improve this answer
    
thanks for your answer, I updated the question accordingly. –  Ewan Delanoy May 10 '12 at 15:33
    
But does recognizing $G$ as a Coxeter group help us with finding all its index $2$ subgroups ? –  Ewan Delanoy May 10 '12 at 15:33
    
@EwanDelanoy: The hint should help you to find a 2nd subgroup of index 2 (which you found). –  j.p. May 10 '12 at 15:56
    
@EwanDelanoy: At least my knowledge about Coxeter groups doesn't help finding all subgroups of index 2. I got this 2nd subgroup by considering $\{\pm 1\}^n$ as $\mathbb{F}_2$ vector space on which $S_n$ acts by permuting a base. As this action is $2$-transitive (for $n\ge 2$), over characteristic $0$ the linear representation would be the sum of two irreducible ones, but not so over characteristic $2$... –  j.p. May 10 '12 at 16:06
    
@EwanDelanoy: I'd guess that there are no other index-2-subgroups of $G$. A possible way to prove this could be to show that all elements of $\{\pm 1\}^n\cap \mathop{Ker}(t')$ are commutators in $G$. –  j.p. May 10 '12 at 16:11
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.