# 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!

• First of all we need $2n\mid \frac{n!}{2}$ by Lagrange. – Dietrich Burde Oct 24 '19 at 18:10
• not a subgroup, but isomorphic to a subgroup! – Hagen von Eitzen Oct 24 '19 at 18:11
• @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 – Hagen von Eitzen Oct 24 '19 at 18:12
• 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..." – Dietrich Burde Oct 24 '19 at 18:29
• 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 $n \equiv 1 \bmod 4$. – Derek Holt Oct 24 '19 at 18:56

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

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 $$A_n$$.

• $D_{2n}$ 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 $n \equiv 1 mod 4$. – jeffery_the_wind Oct 25 '19 at 17:02