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I am trying to identify the following presentation

$\langle x,y : x^{2} , y^{3} , [x,y] , x^{6}y^{6}\rangle$

I substituted the first relation in the final one and got $x^{6}=1$ so the group is cyclic of order 6 The problem is why we can't say that the schreier transversal is $\{x^{0}.....x^{5}\}$ so we omit the other generator $y$.

The correct answer in fact is claiming that the transversal is $\{x^{i}y^{j}\}$. But in another case for The presentation of the cyclic group of order 7 we omit the other generator.

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    $\begingroup$ Have you made a mistake in your relations? The first is $x^2$ and the final is $y^6$, how do you substitute one into the other? $\endgroup$ – Robert Chamberlain May 14 '18 at 14:15
  • $\begingroup$ Yes @Robert Chamberlain $\endgroup$ – Rosa May 14 '18 at 14:21
  • $\begingroup$ I corrected the mistake thanks alot $\endgroup$ – Rosa May 14 '18 at 14:21
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    $\begingroup$ "I substituted the first relation in the final one and got $x^6 =1$". Sure. But this does not imply that the group is cyclic of order 6...($x$ has order 2 so obviously $x^6=1$...I mean, it is also true that $x^{12223534}=1$, but so what?!?) $\endgroup$ – user1729 May 14 '18 at 14:28
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The relators $x^2$, $y^3$ and $[x, y]$ combine to imply the third relator, $x^6y^6$. So your group is: $$ \langle x, y; x^2, y^3, [x, y]\rangle $$ Note that the cross product $\mathbb{Z}_2\times\mathbb{Z}_3$ satisfies this presentation (why?). It then follows that your group is precisely this presentation (why?).

This means that your group is cyclic of order 6 (why? - this is where the transversal being $\{x^iy^j\}$ comes from).

Note that although the group is cyclic of order 6, it is not for the reasons which you give in your question. In particular, in your question you imply that $x$ has order 6 as $x^6=1$. However, $x$ has order 2.

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  • $\begingroup$ So why in the case of cyclic group of order 7 the transversal is simply the cyclic group of order 6 while the the transversal of the giving prisintation needs booth generators @user1729 $\endgroup$ – Rosa May 14 '18 at 15:29
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    $\begingroup$ Well, do you know why $\mathbb{Z}_2\times\mathbb{Z}_3\cong\mathbb{Z}_6$? $\endgroup$ – user1729 May 14 '18 at 15:36
  • $\begingroup$ Yes I know that .my concern is in the transversal @user1729 $\endgroup$ – Rosa May 14 '18 at 15:38
  • $\begingroup$ So you realise that $\mathbb{Z}_6$ can be generated by the elements $2$ and $3$, so $\mathbb{Z}_6=\langle 2, 3\rangle$, even though neither element has order $6$? $\endgroup$ – user1729 May 14 '18 at 17:56
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If you examine $$(xy)^0, (xy)^1, (xy)^2, (xy)^3, (xy)^4, (xy)^5, (xy)^6 \dots$$ you reduce to $$1, xy, y^2, x, y, xy^2, 1 \dots$$ and the sequence repeats. It is just that the presentation you are given does not exhibit a generator of the whole group. There are non-trivial elements of the group such as $x, y, y^2$ which are not generators.

As for the cyclic group of order $7$, since $7$ is prime every non-identity element generates the group.

Is this what is causing your confusion?

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