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# Find all normal subgroups of D2n (dihedral) by treating the cases where n is odd and even separately.?

Consider the conjugacy classes in D2n

Relevance

Terminology:

The dihedral group is generated by a rotation R and a reflection F:

R^n = e

F^2 = e

Elements of the form R^k are "rotations."

Elements of the form R^k F are "reflections."

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Known properties of the dihedral group:

Any element of the dihedral group can be written R^k or R^k F for 0≤k<n.

R^k F = F R^(n-k)

The subgroup generated by R is normal (b.c. it is index 2).

We will let N be the normal subgroup we are trying to track down.

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Let's start. For n=2k+1, the conjugacy classes are:

{e} { R, R^(n-1) } { R^2 , R^(n-2) } ... { R^k , R^(k+1) }

and { R^i F : 0 ≤ i ≤ n }

Recall that an equivalent definition for a normal subgroup is one that is a union of conjugacy classes.

If a single reflection is in N, then they all are. Even worse, that conjugacy class is not a subgroup -- so to contain this class, you need that class and more. That means N is more than half the group, so indeed, that would require N to be the whole group!

So the ONLY normal subgroups that are not all of D2n are contained in <R> the cyclic group generated by R. But you know, this subgroup is isomorphic to Zn. We know the subgroups of that are <R^d>. For each divisor d of n, we get a different normal subgroup.

So the normal subgroups are:

<R^0> = e

<R^d> for all d dividing n, including n.

D2n itself

You can check that, of course, these are unions of conjugacy classes, but we know by other means already explained that these are normal.

And that's all. Pretty simple.

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The even case is, of course, harder. Say n=2k.

First of all, there is actually a center to this group!

Z(D2n) = { e , R^k }.

That is a normal subgroup already that we would have gotten by our previous argument -- since it's inside <R>. So far then, nothing unknown. But what are the conjugacy classes? They are not the same.

They are the following:

{e} { R, R^(n-1) } { R^2 , R^(n-2) } ... { R^(k-1) , R^(k+1) } , { R^k }

and { R^(2i) F : 0 ≤ i ≤ k }, the even reflections

and { R^(2i+1) F : 0 ≤ i ≤ k }, the odd reflections

So we can definitely still get all of the subgroups of <R>.

What else can we get? Well, we have our unknown normal subgroup N. If N contains any reflections, it has at least one quarter of G -- so it's pretty big. Assume it contains R^j F. If j is even, then N contains F and R^2 F, and thus it contains R^2.

If j is odd, then N contains R F and R^3 F, so it contains R^2 = R^3FRF.

So in fact, N contains even more of G -- half of it in fact. It must contain <R^2>, which is a quarter of G, and one of the conjugacy classes -- either the odd reflections, or the even ones. And of course, we don't want to include anything more -- or else N=G.

So in the end, we get the following normal subgroups:

<R^0> = e

<R^d> for all d dividing n, including n.

<R^2,F>

<R^2,RF>

D2n

So in the even case, there are precisely two extra normal subgroups.

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You could also view D2n as the semidirect product of Zn and Z2, where Z2 acts on Zn by the automorphism x → -x (which is an automorphism because Zn is abelian).

From this, you know that you can deal with Zn (which means the subgroup generated by R) has normal subgroups which are obvious (if d divides n, then R^d generates a subgroup of size n/d). Those make up some of them, and the rest are simply those plus any one of the reflections.

In the case where n is odd you are done, like above, but for n even, this doesn't completely help. This very old question:

is an example where hand-waving occurs. It does not account for the possibility that (for example) n = 8. In that case, it is not clear that you can build up the (normal) subgroups that come from <R> which are:

<R>, <R^2> , <R^4>

and join on the other stuff... indeed, it doesn't help because

< R^2 , F > ≠ < R^2 , RF >

even though both of those are normal of index 2.

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