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? 

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