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In a problem from a past exam I am asked "When can $D_n = \langle r,s\mid r^n = s^2 = (rs)^2 = 1\rangle$, the dihedral group of order $2n$, be expressed as a direct product $G\times H$ of two nontrivial groups?" My answer is when $n=2$.

My reasoning is as follows: what I should seek is nontrivial proper normal subgroups $G, H$ of $D_n$ such that $GH=D_n$ and $G\cap H$ is trivial. A nontrivial proper normal subgroup of $D_n$ is cyclic, if $n\in 1+2\mathbb Z$. If $n \in 2\mathbb Z$, subgroups of the form $\langle r^2,s\rangle$ or $\langle r^2, sr\rangle$ are also normal. However, no matter how we choose $G,H$ from these, we cannot satisfy both $GH=D_n$ and $G\cap H = \{1\}$, if $n>2$.

I would be grateful if you could tell me this is correct.

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    $\begingroup$ To get an idea of where you went wrong try and prove that $D_6\cong D_3\times C_2$. Generalize? $\endgroup$ – Jyrki Lahtonen Sep 5 '13 at 16:24
  • $\begingroup$ @JyrkiLahtonen I was reading the condition $GH=D_n$ as $G\cup H=D_n$ and that's why I went wrong. This is my second answer: if $n\in 1+2\mathbb Z$, there is no two normal subgroups satisfying the condition. If $n\in 2\Bbb Z$, one of such normal subgroups (let this be $G$) must contain $s$. Then the order of $G$ is half the order of $D_n$, so the other factor $H$ must be $\langle r^{n/2}\rangle$. Then $GH=D_n$ iff $n/2$ is odd. So the answer is $n$ is even and $n/2$ is odd. Is this ok? $\endgroup$ – Pteromys Sep 7 '13 at 1:31
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I think you make too many assertions which you have not established. Note, for example, that if $n=9$, the subgroup generated by $<r^3,s>$ is normal. Also you have to establish "no matter how we choose $G,H$ from these $\dots$"

I think you would do better to notice that in the direct product $G\times H$ every element of $G$ commutes with every element of $H$. This is a property of the direct product which you have not noted in your attempt. $D$ is not commutative, so this condition puts a restriction on what $G$ and $H$ might be, and the elements they might contain.

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  • $\begingroup$ I read this post: [in.answers.yahoo.com/question/index?qid=20091014113730AA7KJDt] and I was convinced that the subgroups I listed above are the only normal subgroups $D_n$ have. Do you mean its poster and I are wrong? $\endgroup$ – Pteromys Sep 6 '13 at 0:32
  • $\begingroup$ I am not used to Markdown and I made a mistake in posting the link above. Obviously, the link should not contain the last character ]. $\endgroup$ – Pteromys Sep 6 '13 at 0:43

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