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If I use the notation $D_{2n}$, then does $D_4$ make sense?

If I showed that a group $G$ is isomorphic to $H \times D_4$ where $H$ is a group, then is $G$ not a group?

I am asking this because in my other question the answerer didn't directly address my question.

Frenzy Li
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FNH
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    As I see @mrf did it perfect, so why do you think the answer is not satisfying? – Mikasa Jun 10 '13 at 00:25
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    @BabakS. , it proves the statement which my questions asks either my trial does this or it doesn't ! but i think it's a very useful answer though – FNH Jun 10 '13 at 00:34

3 Answers3

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You could interpret $D_4$ as symmetries of a $2$-gon, which has 2 vertices connected by 2 edges. Then swapping the edges is one generator (which has order 2) and swapping the vertices is another generator (which has order 2), so $D_4 \cong Z/2 \times Z/2$ in this case.

Steven Sam
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As $$D_{2n}=\langle x,y\mid x^n=y^2=(xy)^2=1\rangle$$ so $$D_4=\langle x,y\mid x^2=y^2=(xy)^2=1\rangle$$ so $$D_4/\langle x\rangle\cong\mathbb Z_2=\langle y\rangle$$ But $\langle y\rangle$ is normal in $D_4$ so $D_4\cong\mathbb Z_2\times\mathbb Z_2$

Mikasa
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    Hello! and Thank you Amy. :-) – Mikasa Jun 10 '13 at 00:26
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    Hello! Congratulations! $>30$K! – amWhy Jun 10 '13 at 00:27
  • @Babak S. , how is $(y)$ normal affect here ? i mean ,$ |(x)| = 2$ and so is $(y)$ , so let $Z=(x)\times (y)$ , so $|Z|=4$ so $Z \cong D_4$ so $D_4 \cong Z_2 \times Z_2$ and we don't assume $(y)$ to be normal ? of course it's as its index is 2 and the group is abelian , but it doesn't matter as i guess ? $(a)$ means the subgroup generated by $a $ – FNH Jun 10 '13 at 00:29
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    @MathsLover: Yes that is right. It is not affect the problem, however it is normal. – Mikasa Jun 10 '13 at 00:35
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Sure. In that case, $D_{4}$ is the group of symmetries of a $2$-gon, and is in fact isomorphic to the Klein 4-group, $V_{4}$. However, unless I've misunderstood you, I think you ought to think about the answer to your question (the one you linked to) again - it looks perfectly good to me.

Alex Wertheim
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    not quite the symmetries of a line segment (since that would be a group of order $2$). $D_4$ is the group of symmetric of a $2$-gon. A $2$-gon does not exist as a Euclidean polygon, but it exists as an abstract polygon. – Ittay Weiss Jun 10 '13 at 00:39