Show the dihedral group ##D_6## is isomorphic to ##S_3 x Z_2##

[Robb]

Homework Statement

Show that the dihedral group $D_6 is isomorphic to$S_3 x Z_2$by constructing an explicit isomorphism. ( Hint: Color every other vertex of a hexagon red. Which elements of$D_6## permute these 3 vertices. Also, which elements are order 2?)

Relevant Equations

$S_3 = { e, f, g, f^2, gf, gf^2 }$
$Z_2 = { 0,1 }$
$D_6 = { e, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5 }$
$f^3 = e$
$g^2 = e$

I'm not really sure where to begin with this problem. Any insight as to where to begin or what to look out for would be much appreciated!

Orders of $S_3$
$|e|=1$
$|f|=3$
$|f^2|=2$
$|g|=2$
$|gf|=2$
$|gf^2|=3$

Orders of $Z_2$
$|0|=1$
$|1|=2$

Orders of $S_3 x Z_2$
$|e,0|=1$
$|e,1|=2$
$|f,0|=2$
$|f,1|=2$
$|f^2,0|=2$
$|F^2,1|=6$
$|g,0|=2$
$|g,1|=6$
$|gf,0|=3$
$|gf,1|=2$
$|gf^2,0|=2$
$|gf^2,1|=2$

Orders of $D_6$
$|e|=1$
$|r|=6$
$|r^2|=3$
$|r^3|=2$
$|r^4|=3$
$|r^5|=6$
$|s|=2$
$|sr|=2$
$|sr^2|=2$
$|sr^3|=2$
$|sr^4|=2$
$|sr^5|=2$

 

[fresh_42]

What is $D_6$? I'm asking since there are various definitions for it. The coloring hint indicates it is the symmetry group of a hexagon, the representation you gave under Relevant Equations indicates that you can use generators and relations, too.

Furthermore I would write $S_3$ explicitly with permutations rather than with $e,f,g$, but that's only a side note. $S_3=\langle e,f,g\,|\,e=f^3=g^2 \rangle$ works as well. I also would choose only one group operation. As you have chosen the multiplication for $D_6$ and $S_3$, let's do the same with $\mathbb{Z}_2=\{\,-1,+1\,\}$.

Now $S_3 \times \mathbb{Z}_2$ as a set are all pairs $(\sigma,\varepsilon)$ with $\sigma\in S_3$ and $\varepsilon \in \mathbb{Z}_2$. These are $12$ pairs total, same as $12$ elements you listed for $D_6$. Have you tried to establish a correspondence between them? And if possible by keeping the orders of the elements!