For which integers $n \ge 3$ is the dihedral group $D_{2n}$ a sub group of $Alt_n$ 
For which integers $n \ge 3$ is the dihedral group $D_{2n}$ a subgroup of $Alt_n$

Disclaimer, we are supposed to get the answer without Lagrange's Theorem.
I have just started working with the Alternating group $Alt_n$, or also denoted $\mathbb{A}_n$. 
$$Alt_n := \{ f\in S_n \mid \operatorname{sg}(f)=1 \}$$
$\operatorname{sg}(f)=1$ when f is a product even pair of 2-cycles in the symmetric group $S_n$, which equivalently means that f is product of 3-cycles. 
So this is all very new to me, and I am at a loss of how to consider how the dihedral group could be a subgroup of $Alt_n$. Define:
$$D_{2n} = \langle r,s\mid r^n = s^2 = 1, sr^i=r^{-i}s\rangle$$
In this part of the class we are talking about sub groups generated by sub sets of the group; order of subgroups; the subgroup criterion, which says for a subset to be a subgroup then it need to be closed under product and inverses. So I assume the first step is to show that $D_{2n}$ is a sub set of $S_n$. But I'm just not making the connection.
It seems like elements of $S_n$ will always have an odd order, so perhaps $n$ needs to be odd. 
So like I said, I am a little at a loss of even the criteria that we would need to show that $D_{2n}$ is a sub group of $Alt_n$. Thanks for the help!
 A: Idea: We can embed $D_{2n}$ to $S_{n-2}$ for certain $n$, and $S_{n-2}$ can always be embedded into $A_n$ for all $n\ge 2$, see the following posts:
The smallest symmetric group $S_m$ into which a given dihedral group $D_{2n}$ embeds
Embedding $S_n$ into $A_{n+2}$
Hence we can embed $D_{2n}$ into $A_n$ for these $n$. 
We need $n\ge 5$, because for $n=4$, $D_8$ is not isomorphic to a subgroup of $A_4$, because $8\nmid 12$ contradicts Lagrange. If you cannot use Lagrange (as you say), you could use a classification of subgroups of $A_4$ here:
Find the subgroups of A4
A: This additional answer is just to clarify the confusion about what was being asked. Apparently the intended question was, for which $n$ is the subgroup $$\langle (1,2,3,\ldots,n),(2,n)(3,n-1) \cdots \rangle\cong D_{2n}$$ of $S_n$ contained in $A_n$?
For that we need $n$ to be odd, since otherwise the $n$-cycle is an odd permutation. We also need $n \equiv 1 \bmod 4$, since otherwise the reflection $(2,n)(3,n-1) \cdots$ has an odd number of transpositions. But if $n \equiv 1 \bmod 4$ then both of the group generators are even permutations, and so the subgroup lies in $A_n$.
