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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

user1729
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PLAP_
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  • It would be more helpful if you made your question self-contained, rather than requiring the reader to view a different post. Quote the appropriate bits from the linked post, if necessary. – user1729 Oct 03 '18 at 09:28
  • (For your a) it is common to make no distinction between "internal" and "external" direct products, so $\times$ stands for "the direct product". That is, it stands for either/both.) – user1729 Oct 03 '18 at 09:29
  • How can it stand for both? Won't the two require very different approaches? – PLAP_ Oct 03 '18 at 09:58
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    Possibly there are some special cases where they require different approaches, but I am not aware of these. The main thing to bear in mind though is that they define isomorphic groups (that is, if $G_I$ is the internal direct product of $H$ and $K$, and $G_E$ is the external direct product, then $G_I\cong G_E$). – user1729 Oct 03 '18 at 10:29
  • So you mean, either way (assuming IDP or EDP) I will "essentially" get the same end result? – PLAP_ Oct 03 '18 at 10:44
  • Can you please help with the question though? – PLAP_ Oct 03 '18 at 10:44
  • Yes, you "essentially" get the same result. – user1729 Oct 03 '18 at 10:47
  • I would help with question b) but am unwilling to read over another question to do so. – user1729 Oct 03 '18 at 10:48
  • You don't need to - the question reading was only for part a). Now please help. – PLAP_ Oct 03 '18 at 10:50
  • Ah, right - you want to prove that $D_6$ is isomorphic to $S_3 \times \mathbb{Z/2Z}$? Okay. You should make this clearer in your question - put it at the top, and then say "I looked at this question and it made no sense because..." – user1729 Oct 03 '18 at 11:00
  • Please check now. – PLAP_ Oct 03 '18 at 11:05

1 Answers1

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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.
user1729
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  • How are you using D6-S6 and D3-S3 interchangably? I understand for D3 and S3 because of same order. – PLAP_ Oct 03 '18 at 11:15
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    @PLAP_ $S_3$ and $D_3$ are isomorphic; I've added that in to the answer. The $S_6$ is a typo. Fixed now (unless I missed one). – user1729 Oct 03 '18 at 11:20