MHB What are some important properties of the Commutator Group of D_n?

[mathmari]

Hey!

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

I want to show the following:

1. $s^2\in D_n'$
2. $D_n'\cong \mathbb{Z}_n$ if $n$ is odd
3. $D_n'\cong \mathbb{Z}_{\frac{n}{2}}$ if $n$ is even
4. $D_n$ is nilpotent if and only if $n=2^k$ for some $k=1,2,\dots $

I have done the following:

1. $D_n'=\langle [x,y]\mid x,y\in D_n\rangle=\langle xyx^{-1}y^{-1}\mid x,y\in D_n\rangle$
We have that $a,s\in D_n$, so $[s,a]\in D_n' \Rightarrow sas^{-1}a^{-1}\in D_n' \Rightarrow sas^{-1}a\in D_n' \Rightarrow sas^{-1}a\in D_n' \Rightarrow saasa^{-1}a\in D_n' \Rightarrow s^2a^2\in D_n' \Rightarrow s^2\in D_n' $.

Could you give me some hints for 2., 3., 4. ? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

We should figure out what all elements are, starting from the elements $s^k$ and $as^k$. (Thinking)

 

[mathmari]

Hey mathmari! (Smile)

We should figure out what all elements are, starting from the elements $s^k$ and $as^k$. (Thinking)

If $[s^k,as^k]\in D_n'$ we have that $s^kas^ks^{-k}(as^k)^{-1}\in D_n'\Rightarrow s^kas^{-k}a^{-1}\in D_n' \Rightarrow s^kas^{-k}a\in D_n'$.

We have that $asa=s^{-1} \Rightarrow as=s^{-1}a \Rightarrow s^{-k}a=s^{-k+1}(s^{-1}a)=s^{-k+1}as=s^{-k+2}(s^{-1}a)s=s^{-k+2}as^2=\dots =as^k$.

So, $s^kas^{-k}a\in D_n' \Rightarrow s^kaas^k\in D_n' \Rightarrow s^{2k}\in D_n'$, right? (Wondering)

 

[I like Serena]

If $[s^k,as^k]\in D_n'$ we have that $s^kas^ks^{-k}(as^k)^{-1}\in D_n'\Rightarrow s^kas^{-k}a^{-1}\in D_n' \Rightarrow s^kas^{-k}a\in D_n'$.

We have that $asa=s^{-1} \Rightarrow as=s^{-1}a \Rightarrow s^{-k}a=s^{-k+1}(s^{-1}a)=s^{-k+1}as=s^{-k+2}(s^{-1}a)s=s^{-k+2}as^2=\dots =as^k$.

So, $s^kas^{-k}a\in D_n' \Rightarrow s^kaas^k\in D_n' \Rightarrow s^{2k}\in D_n'$, right? (Wondering)

Yep. (Nod)

However, we should evaluate all of $[s^k,s^l]$, $[s^k,as^l]$, $[as^k,s^l]$, and $[as^k,as^l]$.
And then check which additional elements could be generated from those.

Anyway, we will still find the same thing: $D_n' = \{s^{2k} \}$, where $s^n = s^0 = id$.
How many elements do we have if $n$ is even? Or odd? (Wondering)

 

[mathmari]

Anyway, we will still find the same thing: $D_n' = \{s^{2k} \}$, where $s^n = s^0 = id$.
How many elements do we have if $n$ is even? Or odd? (Wondering)

For even $n$ we get the elements of the form $s^{2i}, 1\leq i\leq n$. So since they are only the even powers of $s$, we have $\frac{n}{2}$ elements.

When $n$ is odd we get the following elements:
$s^2, s^4, \dots , s^{2i}, \dots , s^{n-1}, s^1, s^3, \dots , s^{2i+1}, \dots , s^n$
Therefore, we get $n$ elements. Is this correct? (Wondering)

 

[I like Serena]

For even $n$ we get the elements of the form $s^{2i}, 1\leq i\leq n$. So since they are only the even powers of $s$, we have $\frac{n}{2}$ elements.

When $n$ is odd we get the following elements:
$s^2, s^4, \dots , s^{2i}, \dots , s^{n-1}, s^1, s^3, \dots , s^{2i+1}, \dots , s^n$
Therefore, we get $n$ elements.

Is this correct? (Wondering)

Yep. All correct. (Wink)

 

[mathmari]

Yep. All correct. (Wink)

The commutator group $D_n'$ is cyclic, right? (Wondering)

Therefore, we have that $D_n'\cong \mathbb{Z}_n$ if $n$ is odd and $D_n'\cong \mathbb{Z}_{\frac{n}{2}}$ if $n$ is even. For the 4th property do we use the fact that every $p$-group is nilpotent? (Wondering)

 

[mathmari]

If $n=2^k$ then we have that $|D|=2n=2\cdot 2^k=2^{k+1}$.

So, then $D_n$ is a $2$-group.

So, then $D_n$ is nilpotent.
How do we show the other direction? (Wondering)

We suppose that $D_n$ is nilpotent and $n=2^km, \ 2\not\mid m$.

But how do we get a contradiction? (Wondering)