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I already have shown that product $S_3 \times \mathbb{Z/2Z}$ is isomorphic to $S_6$, by taking the subgroups $H = \{id, r^3\}$ and $K = \{id, r^2,r^4,s,sr^2,sr^4\}$ and multiplying $KH$ to get the group $D_6$. I do not know what I can conjecture in general about $D_{2n}$ (or consequently how to prove that).

user1729
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J. Medley
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  • Remember that $S_3 = D_3$ –  Mar 09 '16 at 20:06
  • It isn't a direct product, but it can be isomorphic to one (though I am still not convinced by his proof) –  Mar 09 '16 at 20:16
  • The result you're looking for is an isomorphism $D_{2n} = D_n \times \mathbb{Z}/2\mathbb{Z}$, with certain restrictons on $n$ that I'll let you work out for yourself. To get an idea of what the isomorphism should look like, start by considering the centers of the two groups. – anomaly Mar 09 '16 at 20:32

1 Answers1

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If $n$ is odd, then we have $D_{2n}\simeq D_n\times C_2$. This has been asked before at MSE here. We have $$ D_{2 n} = \langle a, b : a^{2n} = 1, b^{2} = 1, b a b = a^{-1} \rangle. $$ Note that $a^{n}$ has order $2$, and it is central, as it commutes with $a$, and also with $b$, as $b a^{n} b = a^{-n} = a^{n}$. Now we have that $ \langle a^{2}, b \rangle $ is a subgroup of index $2$, isomorphic to $D_{n}$, which intersects $\langle a^{n} \rangle$ trivially.
For $n$ even it is not true. Already $D_4$ is not isomorphic to $D_2\times C_2$, as $D_4$ is not abelian, but $D_2\times C_2$ is.

Community
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Dietrich Burde
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  • To explicitly rule out the decomposition in the case of $n$ even you can observe that $Z(D_{2n})\cong\mathbb{Z}_2$, while $Z(D_n\times C_2)\cong\mathbb{Z}_2\times\mathbb{Z}_2$. – David Hill Mar 09 '16 at 23:00