What I have tried is the following:
Since $D_4$ is a group of order four, then it is abelian, so we use the fundamental theorem of abelian groups which tells us that every finite abelian group is isomorphic to the direct product of cyclic groups. Therefore, $D_4$ has two options. The first is to be isomorphic to a cyclic group of order 4 (but since the elements of $D_4$ do not form a cyclic group, this case is ruled out) or to be isomorphic to the direct product of two cyclic groups of order two, since the other is ruled out possibility, then this is the case.
Assuming my proof was correct, is there another way to prove it?
EDITED:
Considering in $S_Z$ the permutations given by $σ_o(n)=n+1$ and $τ_o(n)=−n$.We define the infinite dihedral group as $D_\infty=<σ_o,τ_o>$. Then we define $D_{2n}=D_\infty/<σ_o^n>$. Considering $σ=σ_oN_n$ and $τ=τ_oN_n$ where $N_n=<σ_o^n>$. Then $D_{2n}=<σ,τ>=\{1,σ,...,σ^{n−1},τ,στ,...σ^{n−1}τ\}$. Considering this definition and my initial question, the value of $n$ will be 2, that is $D_{2*2}=D_4$