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
$$ G = \langle x_1, \ldots , x_k , y_{k+1}, \ldots, y_n \rangle ?$$
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 $G$.
No, $G$ need not equal $\langle x_1, \dotsc, x_k, y_{k + 1}, \dotsc, y_n\rangle$.
Let $G$ be any group generated by two involutions $a, b$. Let $k = 2, n = 4$ and let $x_1 = x_2 = y_3 = y_4 = a$ and $y_1 = y_2 = x_3 = x_4 = b$, so $G = \langle x_i \rangle = \langle y_i \rangle$. We have $x_1x_2 = x_3x_4 = y_1y_2 = y_3y_4$, so $g = 1$ has finite order. Now $\langle x_1, x_2, y_3, y_4 \rangle = \langle x_1 \rangle$ has two elements, so pick any $G$ with more than two elements, for example the Klein four-group.
