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The Problem: Suppose $G=\left\langle x, y, t\mid x^7=y^7=t^3=1, txt^{-1}=x^2, tyt^{-1}=y\right\rangle$. Show that $y\in Z(G)$.

My Attempt: Clearly $y$ commutes with $t$, so $y$ commutes with $t^2$ as well. Thus to show $y\in Z(G)$, it suffices to show that $y$ commutes with $x$. But I struggle to prove that $xy=yx$. It seems a rather silly place to get stuck-any HINT would be greatly appreciated.

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  • $\begingroup$ Why is it that "clearly $y$ commutes with $t$? $\endgroup$
    – coffeemath
    Dec 4, 2022 at 4:19
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    $\begingroup$ @coffeemath $tyt^{-1}=y\implies ty=yt$. $\endgroup$
    – Dick Grayson
    Dec 4, 2022 at 4:21
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    $\begingroup$ It seems like this should be the presentation of a semidirect product of $(\mathbb{Z}/7\mathbb{Z}) * (\mathbb{Z}/7\mathbb{Z})$ by $\mathbb{Z}/3\mathbb{Z}$, so with the two generators of the free product not commuting, they wouldn't commute in the semidirect product either. $\endgroup$
    – Daniel Schepler
    Dec 4, 2022 at 16:13
  • $\begingroup$ @DanielSchepler that's precisely what it is supposed to be. So now-assuming it is indeed the correct presentation-I have to figure out why the center is not trivial since $y$ might not be in there... $\endgroup$
    – Dick Grayson
    Dec 4, 2022 at 17:05

2 Answers 2

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That's not correct. For example. Let $$ x=(1,2,3,4,5,6,7),\ y=(5,8,9,10,11,12,13),\ t=(1,3,4)(2,7,6). $$ It is easy to check that $x^7=y^7=t^3=1$, $txt^{-1}=x^2$, and $tyt^{-1}=y$. However, $xy\neq yx$.

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    $\begingroup$ Nice example, but I think $y$ should begin with a $5$ instead of a $2$--as it is, $y$ and $t$ don't commute. Also $t$ should be inverted if you're multiplying according to the right-to-left convention. $\endgroup$
    – Ravi Fernando
    Dec 4, 2022 at 7:44
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    $\begingroup$ @RaviFernando Thanks, the $2$ is an unfortunate typo, and the multiplication here is as in GAP from left to right. Apparently I should have pointed that out in my answer. $\endgroup$
    – kabenyuk
    Dec 4, 2022 at 8:20
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Further to @kabenyuk's excellent answer, one way to see that this is false is to use GAP:

gap> F:=FreeGroup(3);
<free group on the generators [ f1, f2, f3 ]>
gap> x:=F.1; y:=F.2; t:=F.3;
f1
f2
f3
gap> rels:=[x^7, y^7, t^3, t*x*t^(-1)*x^(-2), t*y*t^(-1)*y^(-1)];
[ f1^7, f2^7, f3^3, f3*f1*f3^-1*f1^-2, f3*f2*f3^-1*f2^-1 ]
gap> G:=F/rels;
<fp group on the generators [ f1, f2, f3 ]>
gap> G.2*G.3=G.3*G.2;  #This is a sanity check: It must return true.
true
gap> G.1*G.2=G.2*G.1;
false
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    $\begingroup$ Too bad I can't accept two different answers... But thank you so much! $\endgroup$
    – Dick Grayson
    Dec 4, 2022 at 15:12

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