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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!

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    $\begingroup$ First of all we need $2n\mid \frac{n!}{2}$ by Lagrange. $\endgroup$ – Dietrich Burde Oct 24 '19 at 18:10
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    $\begingroup$ not a subgroup, but isomorphic to a subgroup! $\endgroup$ – Hagen von Eitzen Oct 24 '19 at 18:11
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    $\begingroup$ @jeffery_the_wind If you didn't have anything by Lagrange yet, you cannot have had much more than the group axioms and I am surprised you have dihedral groups and group presenttion $\endgroup$ – Hagen von Eitzen Oct 24 '19 at 18:12
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    $\begingroup$ I can understand that you are confused. For many readers it sounds a bit strange that you need not use Lagrange. Like you have a question on the zeros of the Riemann zeta function and then you say "in this class they haven't yet introduced prime numbers or anything about primes..." $\endgroup$ – Dietrich Burde Oct 24 '19 at 18:29
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    $\begingroup$ The group $D_{2n}$ is sometimes defined to be the group of rotations and reflections of a regular $n$-gon. So if you thought of it as acting on the vertices, then $D_{2n}$ would be a genuine subgroup of $S_{2n}$. So perhaps the questrion is referring specifically to that subgroup. If so, then the answer is it is contained in $A_n$ iff $n \equiv 1 \bmod 4$. $\endgroup$ – Derek Holt Oct 24 '19 at 18:56
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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

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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 $A_n$.

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  • $\begingroup$ $D_{2n}$ actually turns out to be subgroup of $A_n$ since $A_n := \{ f\in S_n \mid \operatorname{sg}(f)=1 \}$, since for the $n$ that you pointed out $n \equiv 1mod4$, all the elements of $D_{2n}$ can be viewed as permutations $f$ with $sg(f) = 1$ and are in $A_n$ and also $D_{2n}$ is a group (closed under product and inverses), so $D_{2n}$ is a subgroup of $A_n$ when $n \equiv 1 mod 4$. $\endgroup$ – jeffery_the_wind Oct 25 '19 at 17:02

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