Find a nonabelian subgroup $T$ of $S_3 \times \Bbb Z_4$ of index 2, generated by elements $x,y$ such that $|x|=6$, $x^3=y^2$, and $yx=x^{-1}y$.

I gave it a try but still, don't reach the solution.

Consider $H=D_3 \times \Bbb Z_2$, since $D_3 \leq S_3, \Bbb Z_2 \leq \Bbb Z_4$, $H \leq S_3 \times Z_4.$ Consider $x=(\tau,{2}) \in H$ where $\tau=(123)$. Then, we have \begin{align*} x^2=(\tau^2,0),\ \ x^3=(e,2),\ \ x^4=(\tau,0),\ \ x^5=(\tau^2,2),\ \ x^6=(e,0) \end{align*} which is the identity in $G$, so $|x|=6.$ I can not find such a $y$ in my subgroup that satisfies the requirement.

I appreciate any help or hints with that.


Hint: $$\begin{align} S_3 &\cong D_3 \\ &\cong \langle a,b\mid a^3, b^2, bab=a^{-1}\rangle, \end{align}$$

$\Bbb Z_4\cong \langle c\mid c^4\rangle$; and for any groups $G_i\cong\langle X_i\mid R_i\rangle$ for $i\in\{1,2\}$, we have $$G_1\times G_2\cong\langle X_1\cup X_2\mid R_1\cup R_2\cup\{xy=yx\colon x\in X_1, y\in X_2\}\rangle.$$

  • $\begingroup$ What is the meaning of $\langle c\mid c^4\rangle$? $\endgroup$ – manooooh Nov 25 '19 at 0:44
  • 1
    $\begingroup$ It's a presentation of $\Bbb Z_4$, @manooooh. $\endgroup$ – Shaun Nov 25 '19 at 0:45
  • $\begingroup$ Let $G=S_3 \times Z_4$, let $x=(\tau,a) \in G$, $|x|=6$ iff lcm ($|\tau|,|a|$)=6, so we have more than one possibility for that, however, we need such a $y= (\sigma,b) \in G$ such that $x^3=y^2$ i.e $|y|=4=$lcm($|\sigma|,|b|$), so we need $y=(e,b)$ such that $|b|=4$ since $|\sigma|\neq 4$ for all $\sigma \in S_3,$ and there exists an element $b \in Z_4$ such that $|a||4$. The condition $x^3=y^2$ leads to find that $x=((123),2)$, $y=(e,3)$ So $x=((123),2)$, $y=(e,1)$, satisfy the first two conditions. But I still have a problem with $x^-1$!! $\endgroup$ – Ahmed Nov 25 '19 at 4:17

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