Find the subgroup of $\mathbb{Z}_{12}$ generated by the subset $\{4,6\}$. Also draw the digraph of this subgroup $\langle\{4,6\} \rangle$.

Question

I have the following problem:

Find the subgroup of $\mathbb{Z}_{12}$ generated by the subset $\{4,6\}$. Also draw the digraph of this subgroup $\langle\{4,6\} \rangle$.

I've done the first part and attempted the second, but I think I ended up drawing a subgroup diagram instead of a digraph. I'm not sure I understand what a digraph is or how it should look for this problem.

Below is my work:

$\mathbb{Z}_{12}$ is cyclic, so the group generated by $\{4,6\}$ is a cyclic subgroup.

It is known that the subgroup of $\{x_1, x_2, \ldots, x_n\}$ is the same as the subgroup generated by $\gcd (\{ x_1, x_2, \ldots, x_n \})$. So, the subgroup generated by $\{ 4,6 \}$ is the same as the subgroup generated by $\gcd ( \{ 4,6 \}) = 2$.

So, the subgroup required is $\langle 2 \rangle = \{ 0,2,4,6,8,10 \}$.

\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[node distance=2cm]

\title{Digraph of $\langle 2 \rangle$}


 \node(main)        {$e$};
 \node(SGb)      [below left=2cm and 0.6cm of main]  {$\langle 4 \rangle$};
 \node(SGc)      [below right=2cm and 0.6cm of main]       {$\langle 6 \rangle$};
 \node(SGd)      [below =4cm of main]       {$\langle 2 \rangle = \langle \{ 4,6 \} \rangle$};
 \node(ID)            [below = 2cm of SGd]     {$\mathbb{Z}_{12}$};

\draw(main)       -- (SGb);
\draw(main)         -- (SGc);
\draw(SGb)       -- (SGd);
\draw(SGc)       -- (SGd);
\draw(SGd)       -- (ID);


\end{tikzpicture}
\end{center}
\caption{Digraph of $\langle 2 \rangle$}
\end{figure}

.

Have I approached this problem correctly or am I right in my suspicion that I don't really know what a digraph is?

For reference, our textbook refers to these as Cayley Digraphs, I'm not sure that digraph is universally understood.

Should the digraph instead be a hexagon with each vertex an element from {0,2,4,6,8,10} with 2 types of directed edges (1 for 4 and 1 for 6)?

Answer 1

You have approached the question incorrectly; however, your final question indicates that you're almost there.

Cayley Diagrams (as they are also known as) of cyclic (sub)groups are typically drawn as, well, cycles: the vertices are those of a regular $n$-gon, where $n$ is the order of the cyclic (sub)group, and there is one directed edge per adjacent pair of vertices, all going in the same direction. This is due to the fact that these (sub)groups have one generator.

Since $\langle 4,6\rangle=\langle 2\rangle$ is indeed cyclic, the above for $n=\operatorname{ord}_{\Bbb Z_{12}}(2)=6$ applies.

See, say, the beginning of "Combinatorial Group Theory [...]," by Magnus et al.