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

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?

• 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 9 '13 at 21:09
• $D_{2n}$ is the symmetry group of regular $n$-gon,so $|G_{2n}|=2n$ – Fawzy Hegab Jun 9 '13 at 21:13
• 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: math.stackexchange.com/questions/322685 – Watson Aug 29 '16 at 13:34

$D_{8n}$ has an element of order $4n$, but the maximal order of an element in $D_{4n} \times \mathbb{Z}_2$ is $2n$.
• @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? – mrs Jun 10 '13 at 0:44