# Find the subgroups of index two of this finite semi-direct product

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 $5$.

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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 $4$.

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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