A property for a group generated by involutions

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

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

• 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 $g$? – Geoff Robinson Aug 26 '20 at 11:49
• 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. – Stein Chen Sep 2 '20 at 22:48

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 $$G$$.

• 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. – Stein Chen Sep 1 '20 at 18:53

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.

• I had this as a comment and it got dropped out while converting to answer. Will fix later. – Ville Salo Aug 26 '20 at 13:35
• I thank you for that, kind stranger. ("Thanks" was too short.) – Ville Salo Aug 26 '20 at 16:16
• 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.)" :) – Greg Martin Aug 26 '20 at 22:47
• Clever! (I may steal that later, but not immediately.) – Ville Salo Aug 27 '20 at 4:24
• Thank you very much for the answer. – Stein Chen Sep 1 '20 at 18:54