Is there a size 2 generating set of the signed symmetric group $B_n$?

The signed symmetric group $$B_n$$ is a permutation group where the underlying set is $$B_n=\{\sigma \in S_{A_n}| \forall x \in A_n, \sigma(-x)= -\sigma(x)\}$$ with $$A_n=\{-n,-(n-1),-(n-2),\cdots,-1,1,\cdots, n-1,n\}$$.

$$B_n$$ can be generated by the 3 permutations $$(1,2)(-1,-2) ; (1,2,\cdots, n)(-1,-2,\cdots, -n)$$ and $$(-i,i)$$ for some $$i$$.

Is there a generating set of $$B_n$$ with only 2 elements?

• Have you tried to check explicitly for some small values of $n$? – Nate Eldredge Sep 26 '18 at 0:01

Yes. Take two $$a,b$$ generators of $$S_n$$, where $$b$$ has odd order and fixes point $$1$$. For example, if $$n$$ is even, let $$a=(1,2)$$, $$b=(2,3,\ldots,n)$$. If $$n$$ is odd let $$a=(1,2,3,4)$$, $$b=(3,4,\ldots,n)$$.

Let $$\bar{a},\bar{b}$$ be their natural images$$^\dagger$$ in $$B_n$$. Then $$B_n = \langle \bar{a}, \bar{b}(1,-1) \rangle$$. This is because $$\bar{b}$$ and $$(-1,1)$$ commiute and have coprime order, so both $$\bar{b}$$ and $$(-1,1)$$ are powers of $$\bar{b}(-1,1)$$ and hence $$\langle \bar{a}, \bar{b}(1,-1) \rangle=\langle \bar{a}, \bar{b},(1,-1) \rangle =B_n$$.

$$\dagger$$ : For a permutation $$\alpha$$ of $$\{1,2,\ldots,n\}$$ let $$\alpha'$$ be the permutation of $$\{-1,-2,\ldots,-n\}$$ defined by $$\alpha(-i) := -\alpha(i)$$ for $$1 \le i \le n$$. By the natural embeddding $$S_n \to B_n$$ I mean the map defined by $$\alpha \mapsto \alpha\alpha'$$.

• Please what do you mean by natural image? – RTK Sep 27 '18 at 4:14
• $B_n$ is a semidirect product $N \rtimes S_n$, where $|N|=2^n$. This defines a natural embedding $S_n \to B_n$. I have provides it properly in the edited answer. – Derek Holt Sep 27 '18 at 6:54
• If I understand, with the example you provided, when $n$ is even, $\bar{a}=(1,2)(-1,-2)$, $\bar{b}=(2,3,\ldots,n)(-2,-3,\ldots,-n)$. and if $n$ is odd $\bar{a}=(1,2,3,4)(-1,-2,-3,-4)$, $\bar{b}=(3,4,\ldots,n)(-3,-4,\ldots,-n)$. is that correct? – RTK Sep 27 '18 at 17:13
• Yes that is correct – Derek Holt Sep 27 '18 at 18:19
• No it doesn't mean that $B_3$ is not $2$-generated. It just means that you need a different argument. I will leave that to you. – Derek Holt Sep 28 '18 at 21:20

Yes. See proposition 6 in this paper: "The strong symmetric genus of the hyperoctahedral groups" for a recipe of how to find a pair of generators.