4

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.

__

Does this proof make sense or is it mathematically wrong?

FNH
  • 9,130
  • 1
    By $D_{2n}$, you mean the symmetry group of the regular $n$-gon or the regular $2n$-gon? The notation varies a bit. – mrf Jun 09 '13 at 21:09
  • $D_{2n}$ is the symmetry group of regular $n$-gon,so $|G_{2n}|=2n$ – FNH Jun 09 '13 at 21:13
  • 1
    Why do you believe there is no $D_4$? – Steven Stadnicki Jul 20 '15 at 18:42
  • 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. – Steven Stadnicki Jul 20 '15 at 18:45
  • @MathsLover: related: http://math.stackexchange.com/questions/322685 – Watson Aug 29 '16 at 13:34

1 Answers1

7

$D_{8n}$ has an element of order $4n$, but the maximal order of an element in $D_{4n} \times \mathbb{Z}_2$ is $2n$.

mrf
  • 43,639
  • nice , what about my proof ? – FNH Jun 09 '13 at 21:18
  • 2
    @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. – Mikasa Jun 10 '13 at 00:29
  • @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 ! – FNH Jun 10 '13 at 00:32
  • 2
    @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? – Mikasa Jun 10 '13 at 00:44
  • 1
    @BabakS. , you are right! you are so smart Mr Babak :D :D – FNH Jun 10 '13 at 00:48