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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

$$ G = \langle x_1, \ldots , x_k , y_{k+1}, \ldots, y_n \rangle ?$$

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    $\begingroup$ Why would you expect this to be true? Aren't you just saying taht $y_{1} \ldots y_{k} = x_{1} \ldots x_{k}$? And do you mean to assume this for a single $g$? $\endgroup$ – Geoff Robinson Aug 26 '20 at 11:49
  • $\begingroup$ Thank you for the comment. You are right. I forgot to impose some conditions. In the meantime I have written a comment underneath the first answer. $\endgroup$ – Stein Chen Sep 2 '20 at 22:48
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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 $G$.

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    $\begingroup$ Thank you very much for the answer. Despite of the elegant counter-example it would be nice to see if the assertion is correct under some further assumptions: For instance, in the symmetric group, if we only consider transpositions and furthermore the words for $x$ and $y$ are reduced, then the assertion is correct. $\endgroup$ – Stein Chen Sep 1 '20 at 18:53
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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.

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    $\begingroup$ I had this as a comment and it got dropped out while converting to answer. Will fix later. $\endgroup$ – Ville Salo Aug 26 '20 at 13:35
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    $\begingroup$ I thank you for that, kind stranger. ("Thanks" was too short.) $\endgroup$ – Ville Salo Aug 26 '20 at 16:16
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    $\begingroup$ With a mathematician's "sense of humor" I note that you could have also gotten around the minimum length by writing: "Thanks. ('Thanks' was too short.)" :) $\endgroup$ – Greg Martin Aug 26 '20 at 22:47
  • $\begingroup$ Clever! (I may steal that later, but not immediately.) $\endgroup$ – Ville Salo Aug 27 '20 at 4:24
  • $\begingroup$ Thank you very much for the answer. $\endgroup$ – Stein Chen Sep 1 '20 at 18:54

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