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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?

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    $\begingroup$ Have you tried to check explicitly for some small values of $n$? $\endgroup$
    – Nate Eldredge
    Commented Sep 26, 2018 at 0:01

2 Answers 2

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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 $\alpha \mapsto \alpha\alpha'$.

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  • $\begingroup$ Please what do you mean by natural image? $\endgroup$
    – RTK
    Commented Sep 27, 2018 at 4:14
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    $\begingroup$ $B_n$ 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. $\endgroup$
    – Derek Holt
    Commented Sep 27, 2018 at 6:54
  • $\begingroup$ 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? $\endgroup$
    – RTK
    Commented Sep 27, 2018 at 17:13
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    $\begingroup$ Yes that is correct $\endgroup$
    – Derek Holt
    Commented Sep 27, 2018 at 18:19
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    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Commented Sep 28, 2018 at 21:20
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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.

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