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Prove that $D_{12}\cong S_3 \times C_2$.

I really dont know how I should start this question. My gut feeling says in some way I have to consider normal subgroups of $D_{12}$ but I cannot see how this will lead necessarily to a unique solutions.

No full solutions please hints only (partly because I cannot give any more of an attempt and I dont want this downvoted)

Ideally I would like a non geometric solution so that similar techniques can be used for general groups

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  • $\begingroup$ A Cayley table or a Cayley graph is rather large, is there a non-geometrical definition of $D_{12}$ then? $\endgroup$ – Alexey Burdin Apr 27 '15 at 12:10
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    $\begingroup$ There is a more general result in this direction: if $n \geq 6$ is twice an odd number then $D_{2n} \cong D_n \times C_2$. Your question is about $n = 6$. Note $D_6 \cong S_3$ in the notation you are using. $\endgroup$ – KCd Apr 27 '15 at 12:45
  • $\begingroup$ @Permian : related : math.stackexchange.com/questions/322685 $\endgroup$ – Watson Jun 12 '16 at 13:50
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Show that the subgroup generated by $g^3$ is normal; it's obviously isomorphic to $C_2$. Now show that $ D_{12} / C_2 \cong S_3 $.

(Hint: what does the presentation of $D_{12} / C_2$ look like?)

Then look at the subgroup $H$ of $D_{12}$ generated by $\{g^2,h\}$. You can check $H$ has index $2$, and so is normal, and consequently $D_{12}/H \cong C_2$. Then find the structure of $H$.

Or to put it another way, look at the two homomorphisms, one sending $g^3 \mapsto e$, the other sending $g^2,h \mapsto e$.

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    $\begingroup$ It only proves $D_{12}$ is a semi-direct product. $\endgroup$ – Bernard Apr 27 '15 at 12:29
  • $\begingroup$ Why it is semidirect ? Two subgroups are normal. $\endgroup$ – 1ENİGMA1 Dec 15 '17 at 16:20
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You know that $D_{12}$ is the group of symmetries of a Hexagon. Draw the three longest diagonals of the hexagon, and label them $a$, $b$, and $c$. Using this, can you describe a function from $D_{12}$ to $S_3$? What is the kernel of this function? Can you put a copy of $S_3$ back in $D_{12}$?

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    $\begingroup$ (just saw that you want a "non-geometric" solution. What definition of $D_{12}$ do you have in mind?) $\endgroup$ – hunter Apr 27 '15 at 12:07
  • $\begingroup$ $D_{12}=\langle g,h \mid g^6,h^2=1, hgh=g^{-1} \rangle$ $\endgroup$ – Permian Apr 27 '15 at 12:12
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    $\begingroup$ @1234 the copy of $S_3$ is the one generated by $g^2$ and $h$; the copy of $C_2$ is the one generated by $g^3$. Check that the intersection is trivial and both are normal. I have to admit I still "thought geometrically" to see that, though. $\endgroup$ – hunter Apr 27 '15 at 12:19
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    $\begingroup$ What are the three longest diagonals of a hexagon? Especially if it's a regular hexagon :o) $\endgroup$ – Bernard Apr 27 '15 at 12:23
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    $\begingroup$ @Bernard they are the ones going through two opposite points. $\endgroup$ – hunter Apr 27 '15 at 12:25
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The usual well known presentation of dihedral groups gives us in this case:

$$D=\left\{\;s,\,t\;:\;\;s^2=t^6=1\,,\,\,sts=t^5 (=t^{-1})\right\}$$

Now, take

$$\;s:=((12)\,,\,1)\;,\;\;t:=((123)\,,\,c)\in S_3\times C_2\;,\;\;C_2=\{1,c\}\;,\;\;c^2=1$$

Observe that $\;s^2=t^6=1\;$ , and

$$sts=(\,(12)(123)(12)\,,\,1c1\,)=(\,(132),\,\,c\,)$$

and

$$((123)\,,\,c)((132)\,,\,c)=((1)\,,\,1)=1\in S_3\times C_2$$

thus....

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