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.

  • 3
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.