Subgroups of the dihedral group D6

[mathmari]

Hey!

I want to make the diagram for the dihedral group $D_6$:

Subroups of order $2$ : $\langle \tau \rangle$, $\langle \sigma\tau\rangle$, $\langle\sigma^2\tau\rangle$, $\langle\sigma^3\tau\rangle$, $\langle\sigma^4\tau\rangle$, $\langle\sigma^5\tau\rangle$, $\langle\sigma^3\rangle$

Subgroups of order $3$ : $\langle\sigma^2\rangle$, $\langle\sigma^4\rangle$

Subgroups of order $6$ : $\langle \sigma \rangle$, $\langle \sigma^5\rangle$, $\langle\sigma^2, \tau\rangle$, $\langle\sigma^2, \sigma\tau\rangle$

Are there more for each order? (Wondering)

The subgroups of order $4$ are those that are isomorphic to $\mathbb{Z}_4$ or to $\mathbb{Z}_2\times\mathbb{Z}_2$, right? (Wondering)
There are no elements of order $4$, so there are no subgroups of order $\mathbb{Z}_4$, right? (Wondering)
Are the subgroups that are isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ the $\langle a,b\rangle$, for all $a, b$ the elemennt of order $2$
($a,b \in \{\tau, \sigma\tau, \sigma^2\tau , \sigma^3\tau , \sigma^4\tau, \sigma^5\tau , \sigma^3\}$ ) ? (Wondering)

 

[I like Serena]

Hey!

I want to make the diagram for the dihedral group $D_6$:

Subroups of order $2$ : $\langle \tau \rangle$, $\langle \sigma\tau\rangle$, $\langle\sigma^2\tau\rangle$, $\langle\sigma^3\tau\rangle$, $\langle\sigma^4\tau\rangle$, $\langle\sigma^5\tau\rangle$, $\langle\sigma^3\rangle$

Subgroups of order $3$ : $\langle\sigma^2\rangle$, $\langle\sigma^4\rangle$

Subgroups of order $6$ : $\langle \sigma \rangle$, $\langle \sigma^5\rangle$, $\langle\sigma^2, \tau\rangle$, $\langle\sigma^2, \sigma\tau\rangle$

Are there more for each order? (Wondering)

Hey mathmari! (Smile)

I don't think there are more of order 2, 3, and 6.

Oh, and aren't $\langle\sigma^2\rangle$ and $\langle\sigma^4\rangle$ the same sub group?
And $\langle \sigma \rangle$ and $\langle \sigma^5\rangle$ as well? (Wondering)

The subgroups of order $4$ are those that are isomorphic to $\mathbb{Z}_4$ or to $\mathbb{Z}_2\times\mathbb{Z}_2$, right? (Wondering)
There are no elements of order $4$, so there are no subgroups of order $\mathbb{Z}_4$, right? (Wondering)
Are the subgroups that are isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ the $\langle a,b\rangle$, for all $a, b$ the elemennt of order $2$
($a,b \in \{\tau, \sigma\tau, \sigma^2\tau , \sigma^3\tau , \sigma^4\tau, \sigma^5\tau , \sigma^3\}$ ) ? (Wondering)

Yes and yes.
But suppose we pick $a= \tau, b=\sigma\tau$.
Doesn't that mean that $\sigma\tau\cdot \tau = \sigma$ is in the sub group as well?
But then, the whole group is generated!

 

[mathmari]

Oh, and aren't $\langle\sigma^2\rangle$ and $\langle\sigma^4\rangle$ the same sub group?
And $\langle \sigma \rangle$ and $\langle \sigma^5\rangle$ as well? (Wondering)

Ah ok... (Thinking)

But suppose we pick $a= \tau, b=\sigma\tau$.
Doesn't that mean that $\sigma\tau\cdot \tau = \sigma$ is in the sub group as well?
But then, the whole group is generated!

Are the subgroups of order $4$ the following?
$\langle \sigma^3, \sigma\tau\rangle, \langle \sigma^3, \sigma^2\tau\rangle, \langle \sigma^3, \sigma^3\tau\rangle$
(Wondering)
Do you know what command I have to use in Latex to make the diagram? (Wondering)

 

[I like Serena]

Yep.

You mean something like
[TIKZ][-stealth]
\node (id) at (0,0) {$\{1\}$};
\node (s1) at (-2,1) {$\langle\tau\rangle$};
\draw (id) to (s1);
[/TIKZ]
(Wondering)

 

[mathmari]

You mean something like
[TIKZ]
\node (id) at (0,0) {$\{1\}$};
\node (s1) at (-2,1) {$\langle\sigma\rangle$};
\draw (id) to (s1);
[/TIKZ]
(Wondering)

It doesn't appear anything to me:

View attachment 6231

(Wondering)

 

[I like Serena]

It doesn't appear anything to me:

(Wondering)

Something went wrong. I'll have to figure it out when I get home. (Wait)

 

[mathmari]

Something went wrong. I'll have to figure it out when I get home. (Wait)

Ok! (Wait)

 

[mathmari]

Using the following part:

  \node (one) at (0,2) {$\text{\foreignlanguage{english}{id}}$};
  \node (2a) at (-6,0) {$\langle\tau\rangle$};
  \node (2b) at (-4,0) {$\langle\sigma^3\rangle$}; 
  \node (2c) at (-2,0) {$\langle\sigma\tau\rangle$}; 
  \node (2d) at (0,0) {$\langle\sigma^2\tau\rangle$};
  \node (2e) at (2,0) {$\langle\sigma^3\tau\rangle$};
  \node (2f) at (4,0) {$\langle\sigma^4\tau\rangle$};
  \node (2g) at (6,0) {$\langle\sigma^5\tau\rangle$};
  \node (3) at (-5.5,-2) {$\langle\sigma^2\rangle=\langle\sigma^4\rangle$}; 
  \node (4a) at (0.5, -4) {$\langle\sigma^3,\sigma\tau\rangle$}; 
  \node (4b) at (3,-4) {$\langle\sigma^3,\sigma^2\tau\rangle$}; 
  \node (4c) at (5.5,-4) {$\langle\sigma^3,\sigma^3\tau\rangle$}; 
  \node (6a) at (-5, -6.5) {$\langle\sigma\rangle=\langle\sigma^5\rangle$}; 
  \node (6b) at (-2.5, -6.5) {$\langle\sigma^2,\tau\rangle$}; 
  \node (6c) at (3.5, -6.5) {$\langle \sigma^2,\sigma\tau\rangle$}; 
  \node (G) at (0, -8) {$D_6$}; 
  \draw (one) -- (2a) -- (4c) -- (G); 
  \draw (one) -- (2b) -- (4a) -- (G); 
  \draw (2b) -- (4b) -- (G); 
  \draw (2b) -- (4c); 
  \draw (one) -- (2c) -- (4a); 
  \draw (one) -- (2d) -- (4b); 
  \draw (one) -- (2e) -- (4c); 
  \draw (one) -- (2f) -- (4a); 
  \draw (one) -- (2g) -- (4b); 
  \draw (one) -- (3) -- (6a) -- (G); 
  \draw (2b) -- (6a); 
  \draw (3) -- (6b) -- (G); 
  \draw (3) -- (6c) -- (G);

I get the following result:

View attachment 6232

Is this correct? (Wondering)

Or have I forgotten an arrow? (Wondering)

 

[I like Serena]

It works again! (Happy)

\begin{tikzpicture}
%preamble \usepackage{amsmath}
%preamble \usetikzlibrary{shadows}
[very thick,
3dnode/.style 2 args={%
circle,
minimum size=1.2cm,
top color=#1!40!white,
bottom color=#1!60!black,
draw=#1!90!black,
thick,
general shadow={%
shadow xshift=.4ex, shadow yshift=-.4ex,
opacity=.5, fill=black!50,
}
}]
\node

at (-8,2) {Order \ 1};
\node[3dnode={orange}{}] (one) at (0,2) {$\text{id}$};
\node

at (-8,0) {Order \ 2};
\node[3dnode={green}{}] (2a) at (-6,0) {$\langle\tau\rangle$};
\node[3dnode={green}{}] (2b) at (-4,0) {$\langle\sigma^3\rangle$};
\node[3dnode={green}{}] (2c) at (-2,0) {$\langle\sigma\tau\rangle$};
\node[3dnode={green}{}] (2d) at (0,0) {$\langle\sigma^2\tau\rangle$};
\node[3dnode={green}{}] (2e) at (2,0) {$\langle\sigma^3\tau\rangle$};
\node[3dnode={green}{}] (2f) at (4,0) {$\langle\sigma^4\tau\rangle$};
\node[3dnode={green}{}] (2g) at (6,0) {$\langle\sigma^5\tau\rangle$};
\node

at (-8,-2) {Order \ 3};
\node[3dnode={green}{}] (3) at (-5.5,-2) {$\langle\sigma^2\rangle=\langle\sigma^4\rangle$};
\node

at (-8,-4) {Order \ 4};
\node[3dnode={green}{}] (4a) at (0.5, -4) {$\langle\sigma^3,\sigma\tau\rangle$};
\node[3dnode={green}{}] (4b) at (3,-4) {$\langle\sigma^3,\sigma^2\tau\rangle$};
\node[3dnode={green}{}] (4c) at (5.5,-4) {$\langle\sigma^3,\sigma^3\tau\rangle$};
\node

at (-8,-6.5) {Order \ 6};
\node[3dnode={green}{}] (6a) at (-5, -6.5) {$\langle\sigma\rangle=\langle\sigma^5\rangle$};
\node[3dnode={green}{}] (6b) at (-2.5, -6.5) {$\langle\sigma^2,\tau\rangle$};
\node[3dnode={green}{}] (6c) at (3.5, -6.5) {$\langle \sigma^2,\sigma\tau\rangle$};
\node

at (-8,-8) {Order 12};
\node[3dnode={blue}{},text=white] (G) at (0, -8) {$D_6$};
\draw (one) -- (2a) -- (4c) -- (G);
\draw (one) -- (2b) -- (4a) -- (G);
\draw (2b) -- (4b) -- (G);
\draw (2b) -- (4c);
\draw (one) -- (2c) -- (4a);
\draw (one) -- (2d) -- (4b);
\draw (one) -- (2e) -- (4c);
\draw (one) -- (2f) -- (4a);
\draw (one) -- (2g) -- (4b);
\draw (one) -- (3) -- (6a) -- (G);
\draw (2b) -- (6a);
\draw (3) -- (6b) -- (G);
\draw (3) -- (6c) -- (G);
\end{tikzpicture}

Shouldn't there be a couple more arrows to the subgroups of order 6? (Wondering)​

 

[I like Serena]

Shouldn't there be an arrow from $\langle\tau\rangle$ to $\langle \sigma^2,\tau\rangle$? (Wondering)