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 $yx=x^{-1}y$.

Question

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.

Answer 1

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 $$G_1\times G_2\cong\langle X_1\cup X_2\mid R_1\cup R_2\cup\{xy=yx\colon x\in X_1, y\in X_2\}\rangle.$$

Answer 2

Let $x=((123),2), y=((23),1)$, verify that $|x|=6, x^3=y^2=(e,2)$ and $yx=x^{-1}y$. We now show that the subgroup $T\mathrel{\mathop:}=\langle x,y\rangle$ of $S_3 \times \Bbb Z_4$ generated by $x$ and $y$ is a nonabelian group of order $12$. The subgroup $T$ is nonabelian because $yx\not = xy$. The elements of $T$ are products of finitely many $x$'s and $y$'s. Using the equality $yx=x^{-1}y$, every product of finitely many $x$'s and $y$'s can be rewritten so that all $x$'s precede all $y$'s. Since $|x|=6, x^3=y^2$, every element of $T$ is a product of fewer than $6$ $x$'s followed by fewer than $2$ $y$'s; in other words, $T=\{x^iy^j : 0\leq i<6,0\leq j<2\}$. In particular, $T$ has at most $12$ elements. It remains to show that the elements $x^iy^j$ are distinct: Consider the elements $x^ay^b$ and $x^cy^d$ with $0\leq a,c<6,0\leq b,d<2$ and $(a,b)\not=(c,d)$. If $b=d$, then $a\not=c$ and hence $x^ay^b\not=x^cy^d$. If $b\not=d$, or without loss of generality $b=0,d=1$, then the second component of $x^ay^b$ is $0$ or $2$, while the second component of $x^cy^d$ is $1$ or $3$, so $x^ay^b\not=x^cy^d$.