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I was able to find an element of order 6 within $S_3 \times \mathbb{Z}_2$. I was also able to define a function mapping all 6 elements of $S_3 \times \mathbb{Z}_2$ to $D_6$ I the homomorphism portion would be too exhaustive to prove all 36 cases. How do I prove this.

user372382
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  • There's something wrong with your second sentence. Can you give the normal subgroups of $D_6$ isomorphic to $S_3$ and $\Bbb Z_2$ explicitly and write down the homomorphism $S_3\times\Bbb Z_2\to D_6$? – Ted Shifrin Apr 22 '18 at 20:55
  • @TedShifrin Good catch, I meant to write $S_3 \times \mathbb{Z}_2$. I was able to created a function by explicitly assigning elements of $S_3 \times \mathbb{Z}_2$ to $D_6$. It appears to be a homomorphism as well as a bijection but an exhaustive proof would be very tedious. – user372382 Apr 22 '18 at 21:01
  • I don't know what theorems you have at your disposal, but there are some useful ones for such questions. For example, if $H$ and $K$ are normal subgroups of $G$ so that $H\cap K = {e}$ and every element of $G$ can be written as a product of an element of $H$ and an element of $K$, then $G\cong H\times K$. You could probably proudly write a proof of this. :) – Ted Shifrin Apr 22 '18 at 21:04
  • @TedShifrin Thank you for that insight. So essentially what this statement is saying is that if you can create a partition of the set into two normal subgroups then you can reconstruct the entire extended group from their products. – user372382 Apr 22 '18 at 21:14
  • No, you're confusing hypothesis and conclusion, i think. We need to know that we can obtain every element (uniquely) as a product. Then the direct product isomorphism is a consequence. – Ted Shifrin Apr 22 '18 at 21:17
  • Well, it so happens that for any odd $n$, $D_{n}\times \Bbb Z_2 \cong D_{2n}$. It is also the case that $D_3\cong S_3$. I am of course using the notation such that $|D_n|=2n$. – CogitoErgoCogitoSum Apr 22 '18 at 22:54

1 Answers1

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The best way to find an isomorphism is to picture the sets that these groups are naturally acting on. $D_6$ consists of the symmetries of a rgular hexagon, while $S_3\cong D_3$ is the symmetries of an equilateral triangle. This suggests that the $S_3$ component of $S_3\times Z_2$ should be acting on some triangle embedded in the hexagon.

There are only two such triangles which are subsets of the hexagon, and these are $180^\circ$ rotations of each other. Furthermore, $180^\circ$ roation commutes with every symmetry of $D_6$, and $180^\circ$ rotation behaves like $Z_2$. This gives you your isomorphism: $S_3$ acts on the hexagon by rotations/reflections which preserve the two equilateral subtriangles, namely, rotation by multiples of $120^\circ$ and reflection through these triangles' axes of symmetry, while $Z_2$ is $180^\circ$ rotation.

As for the explicit mapping, note that $S_3$ is generated by $(1\,2)$ and $(1\,2\,3)$, $Z_2$ is generated by $z$, and $D_3$ is generated by $r$ and $s$, corresponding to rotation by $60^\circ$ and reflection through some axis. The isomorphism I described is given by $$ (1\,2)\mapsto s,\qquad (1\,2\,3)\mapsto r^2,\qquad z\mapsto r^3 $$ The same idea works for any even polygon to show that $D_{2n}\cong D_n\times Z_2$.

Mike Earnest
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  • That is a very interesting way to visualize it thank you for giving me a deeper intuition into the problem. I am curious then could you do this with any polygon with an even number of sides. More pertinent how would you create a function to express these two sets of mappings to $S_3$ – user372382 Apr 22 '18 at 21:12
  • See the edit $,$ – Mike Earnest Apr 22 '18 at 21:23