4

Prove that $S_3\times \mathbb{Z}_2$ is isomorphic to $D_6$.

My attempt:

I tried to decompose $D_6$ as the internal direct product of subgroups $H$ and $K$ isomorphic to $S_3$ and $\mathbb{Z}_2$ respectively:

Let $H=\{1, r^2, r^4, s, r^2s, r^4s\}$

Let $K=\{1, r^3s\}$

$HK=D_6$

$H\cap K=\{1\}$

However, taking $h=r^2$ and $k=r^3s$, \begin{equation*} \begin{split} hk&=r^2\cdot r^3s \\ &=r^5s \end{split} \end{equation*} \begin{equation*} \begin{split} kh&=r^3s\cdot r^2 \\ &=r^3sr^2ss \\ &=r^3r^{-2}s \\ &= rs \end{split} \end{equation*} $hk\neq kh$

So, this doesn't work. How do I solve this problem?

Siddhartha
  • 1,770
  • 18
  • 38

1 Answers1

5

Note that $Z(D_6) = \langle r^3\rangle$. This is your candidate for the copy of $C_2$.

Pedro
  • 122,002