Let $G$ be a group and $x\_1,\ldots,x\_n,y\_1,\ldots,y\_n \in G$ involutions
such that


* $G = \langle x\_1, \ldots , x\_n \rangle = \langle y\_1 , \ldots ,
y\_n \rangle$
* $g:=x\_1 \cdots x\_n = y\_1 \cdots y\_n$ is of finite order


Now assume that there exists $1 \leq k < n$ such that


$$ g = x\_1 \cdots x\_k y\_{k+1} \cdots y\_n.$$


Is it true that


$$ G = \langle x\_1, \ldots , x\_k , y\_{k+1}, \ldots, y\_n \rangle ?$$



A property for a group generated by involutions
No, this is false even when $G$ is abelian and finite. For instance take


$$G = \langle (1,2), (3,4), (5,6), (7,8) \rangle \le \mathrm{Sym}\_8.$$
Define $x\_1,x\_2,x\_3,x\_4,y\_1,y\_2,y\_3,y\_4$ as indicated by


$$G = \bigl\langle (1,2)(5,6), (3,4)(5,6), (5,6), (7,8) \bigr\rangle. $$


and


$$G = \bigl\langle (1,2), (3,4), (1,2)(5,6), (1,2)(7,8) \bigr\rangle $$


The products of generators are


$$x\_1x\_2x\_3x\_4 = y\_1y\_2y\_3y\_4 = (1,2)(3,4)(5,6)(7,8),$$


which is $x\_1x\_2y\_3y\_4 = (1,2)(5,6)\:(3,4)(5,6)\:(1,2)(5,6)\:(1,2)(7,8)$. But


$$\bigl\langle x\_1,x\_2,y\_3,y\_4\bigr\rangle = \bigl\langle (1,2)(5,6), (3,4)(5,6), (1,2)(5,6), (1,2)(7,8) \bigr\rangle$$


has index $2$ in $G$.


