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?



Is there a size 2 generating set of the signed symmetric group $B_n$?
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 $\alpha \mapsto \alpha\alpha'$.


