Establishing isomorphism between the given two groups

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The question essentially is: prove $D_6$ is isomorphic to $S_3 \times \mathbb{Z/2Z}$.

To me the linked question doesn't make sense because it seems more like trial and error, by finding 2 Normal subgroups $H$ and $K$ of $D_6$ such that $D_6 = H \times K$, and then using the theorem that IDP is isomorphic to EDP. But how in the first place were $H$ and $K$ contructed?

So I have the following questions:

a) $S_3 \times \Bbb Z/2 \Bbb Z$ stands for internal direct product or external direct product? (Apparently Gallian uses $\times$ for IDP only).

b) I am still not able to get how the isomorphism was established.

EDIT: Updated as per requests.

Thanks

Answer 1

For the question about direct products: $H\times K$ is used for "the direct product". The internal and the external direct products of the groups $H$ and $K$ are isomorphic, and we just denote this by $H\times K$. That is, in practice it is common to make no distinction between the internal and the external direct products.

For the more involved question of proving that $D_6\cong S_3\times\mathbb{Z}_2$, some hints. But first, to address the question on finding the subgroups, the main step is step (1). I know from memory/experience/whatever that dihedral groups of $2n$-gons have centre of order two. This was my starting point. Perhaps a different starting point would have been to start with step (2) and to play about with the natural copy of $S_3$ which you have.

1. The dihedral group $D_6$ has centre $Z(D_6)$ of order two. Find the element which generates it. Lets call this element $g$.
2. Note that there exists a natural copy $H$ of $D_3$ in $D_6$ - rotate your hexagon by $2$ steps each time.
3. Note that $g\not\in H$. Hence, $Z(D_6)\cap H=\{1\}$.
4. Note that $H$ has index two in $D_6$.
5. Use (3) and (4) to conclude that $D_6=HD_3$.
6. Use (4) to conclude that $H\lhd D_6$.
7. Conclude that $D_6$ is the internal direct product of $H$ with $Z(D_6)$.
8. Prove that $D_3\cong S_3$. Fin.