Subgroups of index 2 of $\mathbb{Z}_2 \times S_3$

Question

I know that the group $G=\mathbb{Z}_2 \times S_3$ has $3$ subgroups of index $2$ (i.e. of $6$ elements). It is easy to see that $2$ of them are $\{0\} \times S_3$ and $\mathbb{Z}_2 \times A_3$, but I am struggling to find the third one, that probably isn't a direct product of a subgroup of $\mathbb{Z}_2$ and a subgroup of $S_3$ like the first $2$. What can I do?

Answer 1

The third subgroup is a "diagonal". It consists of all elements of the form $$ \{(\operatorname{sgn}\sigma, \sigma)\mid \sigma\in\Bbb S_3\} $$ Each of these subgroups appear as the kernel of a surjective homomorphism onto $\Bbb Z_2$ (index 2 subgroups always do). This probably would've helped you find the final subgroup.

Answer 2

Hint: We know that $\Bbb Z/2\times S_3\cong D_6$, and we know all subgroups of dihedral groups. Then we just pick out the ones of order $6$.

See for example How many subgroups of order $n$ does $D_n$ have?