How can we find all the normal subgroups?

[mathmari]

Hey!

I want to find all the normal subgroups of $D_n$.

We have that $$D_n=\langle a, s \mid a^2=1=s^n, asa=s^{-1}\rangle$$

We have that $K$ is a normal subgroup of $D_n$ iff $$gkg^{-1}\in K, \forall g\in D_n \text{ and } \forall k\in K$$

Could you give me some hints how we could find all these subgroups? (Wondering)

 

[I like Serena]

Hey!

I want to find all the normal subgroups of $D_n$.

We have that $$D_n=\langle a, s \mid a^2=1=s^n, asa=s^{-1}\rangle$$

We have that $K$ is a normal subgroup of $D_n$ iff $$gkg^{-1}\in K, \forall g\in D_n \text{ and } \forall k\in K$$

Could you give me some hints how we could find all these subgroups? (Wondering)

Hey mathmari! (Smile)

How about enumerating all subgroups and see where that brings us? (Wondering)
To do so, we need to pick 1 or 2 (or more) elements and see which group it generates, and then see what it takes to make it normal.

 

[mathmari]

How about enumerating all subgroups and see where that brings us? (Wondering)
To do so, we need to pick 1 or 2 (or more) elements and see which group it generates, and then see what it takes to make it normal.

Do we maybe pick the elements $as^i$ and $s^j$ for $0\leq i,j <n$ ? (Wondering)

 

[Deveno]

Try finding all the cyclic subgroups, first.

With dihedral groups it is helpful to remember there are two types of elements: reflections, and rotations. The reflections have an "$a$" in them.

For subgroups generated by two elements, we have three main kinds of subgroups:

1. Generated by two rotations. This type should be trivial to analyze.
2. Generated by a reflection, and a rotation. This should be easy once you see what happens:

Hint: $s^ka = as^{-k}$, for all $k \in \Bbb N$. Use induction to prove this.

3. Generated by two distinct reflections. This one is a bit tricky.

General warning: it matters whether or not $n$ is even or odd (the even $n$'s produce more subgroups). The prime factorization of $n$ also matters.

 

[mathmari]

Try finding all the cyclic subgroups, first.

Is $\langle s^i\rangle$ for $i\mid n$ a subgroup of $D_n$ ? (Wondering)

If it is, do we have to check if an element of the form $s^ka$ belongs to this subgroup or if it belongs to or generate an other subgroup? (Wondering)

For subgroups generated by two elements, we have three main kinds of subgroups:

1. Generated by two rotations. This type should be trivial to analyze.
2. Generated by a reflection, and a rotation. This should be easy once you see what happens:
3. Generated by two distinct reflections. This one is a bit tricky.

1. Is it generated by a "$s$" ?
2. Is this subgroup of the form $\langle a, s\rangle$ ?
3. Is it generated by an "$a$" ?

(Wondering)

 

[johng1]

I think you're working too hard. As you showed previously, any subgroup of $<s>$ is normal since $<s>$ is normal. So the normal subgroups are the identity, $D_n$, any subgroup of $<s>$ and maybe some more. So suppose $H$ is a proper normal subgroup that is not a subgroup of $<s>$. Then $H$ contains an element of the form $s^ka$ for some $k$. Compute the conjugate $$s^{-1}s^kas=s^ks^{-1}as=s^ka(as^{-1}a)s=s^kas^2\in H$$
So $s^2\in H$. If $n$ is odd, $K=<s^2>=<s>\subset H$ and so $H=D_n$. Suppose $n$ is even. Consider the factor group $D_n/K$. Easily then $H/K$ is one of the 3 proper normal subgroups of $D_n/K$. Thus $H=<s^2,a>$ or $H=<s^2,sa>$, each of these is isomorphic to $D_m$ where $m=n/2$.

 

[mathmari]

So suppose $H$ is a proper normal subgroup that is not a subgroup of $<s>$. Then $H$ contains an element of the form $s^ka$ for some $k$. Compute the conjugate $$s^{-1}s^kas=s^ks^{-1}as=s^ka(as^{-1}a)s=s^kas^2\in H$$
So $s^2\in H$.

Why does this stand? Why does $H$ contain an element of the form $s^ka$ ? (Wondering),

How do we conclude that $s^kas^2\in H$ ? (Wondering)

 

[johng1]

$D_n=\{s^k\,:\,0\leq k<n\}\cup\{s^ka\,:\,0\leq k<n\}$. So if $H$ is not a subgroup (subset) of $<s>$, H contains an element of the form $s^ka$.

By definition of normality of $H$, any conjugate of an element of $H$ is also an element of $H$. So $s^kas^2\in H$ implies $s^2=(s^ka)^{-1}s^kas^2\in H$

 

[mathmari]

Let $H$ be a normal subgroup of $D_n$.

1. If all the elements of $H$ are of the form $s^k$ then $H=\langle s^d\rangle$ for some $d$ that divides $n$.
   
   $\langle s^d\rangle$ is indeed a normal subgroup of $D_n$:
   It suffices to show that $gs^dg^{-1}\in \langle s^d\rangle, \forall g\in D_n$.
   • if $g=s^k$, then $s^ks^ds^{-k}=s^{k+d-k}=s^d\in \langle s^d\rangle$
     
   • if $g=as^k$, then $as^ks^d(as^k)^{-1}=as^ks^ds^{-k}a^{-1}=as^ks^ds^{-k}a=as^{k+d-k}a=as^da=a\underset{d \text{ times } "s"}{\underbrace{saasaas\dots as}a}=(asa)^d=(s^{-1})^d=s^{-d}\in \langle s^d\rangle$
2. If $H$ contains an element of the form $s^ka$, for some $k$, then since it is a normal subgroup of $D_n$, it must hold that $$s^{-1}s^kas\in H \Rightarrow s^{-1+k}as\in \Rightarrow s^ks^{-1}as\in H \Rightarrow s^kas^2\in H$$
   
   So $s^2=(s^ka)^{-1}s^kas^2\in H$.
   
   But how can we find exactly which this $H$ is? (Wondering)

 

[johng1]

I'm not exactly sure what your question is. Maybe this is what you're asking:
Assume $s^ka\in H$. If $k$ is even, say $k=2j$, then since $<s^2>\subseteq H$, $a=s^{-2j}s^{2j}a\in H$ and so $H=<s^2,a>$. Similarly, if $k$ is odd, $H=<s^2,sa>$.

 

[mathmari]

Assume $s^ka\in H$. If $k$ is even, say $k=2j$, then since $<s^2>\subseteq H$, $a=s^{-2j}s^{2j}a\in H$ and so $H=<s^2,a>$.

Why does it stand that $<s^2>\subseteq H$ ? (Wondering)

Having shown that $s^ka\in H$ and $a\in H$, why does it imply that $H=<s^2,a>$ ? (Wondering)

How do we know that these two elements generate a subgroup? (Wondering)

Similarly, if $k$ is odd, $H=<s^2,sa>$.

Assume $s^ka\in H$. When $k$ is odd, say $k=2j+1$, does it stand that $<s^2>\subseteq H$ ? (Wondering)
If it holds, then $sa=ss^{-2j-1}s^{2j+1}a=s^{-2j}s^{2j+1}a\in H$ and so $H=<s^2,sa>$, right? (Wondering)

 

[mathmari]

$$D_n=\langle a,s\mid s^n=1=a^2, asa=s^{-1}\rangle$$

Let $N=\langle s\rangle=\{s, s^2, \dots, s^{n-1}, s^n=1\}$.
We have that $[D_n:N]=\frac{|D_n|}{|N|}=\frac{2n}{n}=2$.
So $N$ is a normal subgroup of $D_n$. Therefore, each subgroup of $N$ is a normal subgroup of $D_n$. (As shown in the thread http://mathhelpboards.com/linear-abstract-algebra-14/normal-subgroup-g-17852.html )

Let $T\neq N$, be a normal subgroup of $D_n$.

So, $\exists \ as^k$, for some $k=1, 2, \dots , n$ with $$s^{-i}(as^k)s^i\in T , \ \forall i=1, 2, \dots , n$$

Since $asa=s^{-1} \Rightarrow as=s^{-1}a \Rightarrow s^{-i}a=s^{-i+1}(s^{-1}a)=s^{-i+1}as=s^{-i+2}(s^{-1}a)s=s^{-i+2}as^2=\dots =as^i$, we have that $$s^{-i}as^ks^i\in T , \ \forall i=1, 2, \dots , n \\ \Rightarrow as^is^ks^i\in T , \ \forall i=1, 2, \dots , n \\ \Rightarrow as^{k+2i}\in T , \ \forall i=1, 2, \dots , n$$

We also have that $$a(as^k)a\in T \Rightarrow s^ka \in T\Rightarrow as^{-k}\in T$$

Then the product of these is also an element of $T$, $as^{-k}as^{k+2i}\in T \Rightarrow aas^ks^{k+2i}=s^{2k+2i}=(s^2)^{k+i}\in T$.

Now we have to take cases if $n$ is odd or even, or not? (Wondering)

If $k$ is even, say $k=2j$, we have that $as^{k+2i}\in T \Rightarrow as^{2j+2i}\in T\Rightarrow a(s^2)^{j+i}\in T\Rightarrow a\in T$.
If $k$ is odd, say $k=2j+1$, we have that $as^{k+2i}\in T \Rightarrow as^{2j+1+2i}\in T\Rightarrow as(s^2)^{j+i}\in T\Rightarrow as\in T$.
Does these two cases hold in both cases, when $n$ is odd and even? (Wondering)