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Prove $D_{8n} \not\cong D_{4n} \times Z_2$.

My trial:

I tried to show that $D_{16}$ is not isomorphic to $D_8 \times Z_2$ by making a contradiction as follows:

Suppose $D_{4n}$ is isomorphic to $D_{2n} \times Z_2$, so $D_{8}$ is isomorphic to $D_{4} \times Z_2$. If $D_{16}$ is isomorphic to $D_{8} \times Z_2 $, then $D_{16}$ is isomorphic to $D_{4} \times Z_2 \times Z_2 $, but there is not Dihedral group of order $4$ so $D_4$ is not a group and so $D_{16}\not\cong D_8\times Z_2$, which gives us a contradiction. Hence, $D_{16}$ is not isomorphic to $D_{8} \times Z_2$.

I found a counterexample for the statement, so it's not true in general, or at least it's not true in this case.

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Does this proof make sense or is it mathematically wrong?

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    $\begingroup$ By $D_{2n}$, you mean the symmetry group of the regular $n$-gon or the regular $2n$-gon? The notation varies a bit. $\endgroup$ – mrf Jun 9 '13 at 21:09
  • $\begingroup$ $D_{2n}$ is the symmetry group of regular $n$-gon,so $|G_{2n}|=2n$ $\endgroup$ – Fawzy Hegab Jun 9 '13 at 21:13
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    $\begingroup$ Why do you believe there is no $D_4$? $\endgroup$ – Steven Stadnicki Jul 20 '15 at 18:42
  • $\begingroup$ Also, there are two distinct possibilities here: one is 'prove that it is not true that for all $n$, $D_{8n}\cong D_{4n}\times Z_2$'; the other is 'prove that for all $n$ it is not the case that $D_{8n}\cong D_{4n}\times Z_2$'. Your argument is trying to show the former, but the exercise is almost certainly asking you to prove the latter. $\endgroup$ – Steven Stadnicki Jul 20 '15 at 18:45
  • $\begingroup$ @MathsLover: related: math.stackexchange.com/questions/322685 $\endgroup$ – Watson Aug 29 '16 at 13:34
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$D_{8n}$ has an element of order $4n$, but the maximal order of an element in $D_{4n} \times \mathbb{Z}_2$ is $2n$.

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  • $\begingroup$ nice , what about my proof ? $\endgroup$ – Fawzy Hegab Jun 9 '13 at 21:18
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    $\begingroup$ @MathsLover: As you formed the question, you need just an inconsistency in two groups and this answer brought you a contradiction in a very nice way. $\endgroup$ – mrs Jun 10 '13 at 0:29
  • $\begingroup$ @BabakS. , yes , it brought me a contradiction but my question was if my way of solving the exercise brings this contradiction or not ! the question which you have answered minutes ago shows that my trial is completely wrong ! $\endgroup$ – Fawzy Hegab Jun 10 '13 at 0:32
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    $\begingroup$ @MathsLover: $D_4$ is abelian, so $D_4\times\textbf{V}\cong D_{16}$ is belian which is wrong so there should be an defect in that. right? $\endgroup$ – mrs Jun 10 '13 at 0:44
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    $\begingroup$ @BabakS. , you are right! you are so smart Mr Babak :D :D $\endgroup$ – Fawzy Hegab Jun 10 '13 at 0:48

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