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problem:aut-square [2013/08/09 06:35] joshuawiscons created |
problem:aut-square [2013/08/22 16:30] (current) bmwoodruff |
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====== Automorphism group of a square ====== | ====== Automorphism group of a square ====== | ||
- | **Problem.** Consider the graph $\mathcal{G} = (V,E)$ drawn below. The vertex set is $V$ and the (symmetric) relation giving adjacency is $E$. Specifically, | + | |
- | Write down all elements of $\textrm{Aut}(\mathcal{G})$; | + | ==== Problem |
- | - There some $a\in \textrm{Aut}(\mathcal{G})$ such that $a^2$ is the identity. | + | Consider the graph $\mathcal{G} = (V,E)$ drawn below. The vertex set is $V$ and the (symmetric) relation giving adjacency is $E$. Specifically, |
- | - There some $a\in \textrm{Aut}(\mathcal{G})$ such that $a^3$ is the identity. | + | Write down all elements of $\aut(\mathcal{G})$; |
- | - There some $a\in \textrm{Aut}(\mathcal{G})$ such that every other $b\in \textrm{Aut}(\mathcal{G})$ is of the form $b=a^n$ for some $n$. | + | - There is an $a\in \aut(\mathcal{G})$ such that $a^2$ is the identity but $a$ is not the identity. |
- | - For every $a,b\in \textrm{Aut}(\mathcal{G})$, | + | - There is an $a\in \aut(\mathcal{G})$ such that $a^3$ is the identity but $a$ is not the identity. |
+ | - There is an $a\in \aut(\mathcal{G})$ such that every other $b\in \aut(\mathcal{G})$ is of the form $b=a^n$ for some $n$. | ||
+ | - For every $a,b\in \aut(\mathcal{G})$, | ||
{{ wiki: | {{ wiki: | ||
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---- | ---- | ||
==== Remarks ==== | ==== Remarks ==== | ||
- | * $\textrm{Aut}(\mathcal{G})$ denotes the [[definition: | + | * $\aut(\mathcal{G})$ denotes the [[definition: |
* $S_4$ denotes the [[definition: | * $S_4$ denotes the [[definition: | ||
---- | ---- | ||
==== $\LaTeX$ version ==== | ==== $\LaTeX$ version ==== | ||
- | <code> | + | <file tex problem.aut-square.tex> |
- | %%%%%%%%%% | + | %%%%% |
% DEPENDENCIES | % DEPENDENCIES | ||
- | % RequiredPackages: \usepackage{tikz} | + | % RequiredPackages \usepackage{tikz} |
- | % RequiredMacros: | + | % RequiredMacros: |
- | %%%%%%%%%% | + | %%%%% |
\begin{problem} | \begin{problem} | ||
- | Consider the graph $\mathcal{G} = (V,E)$ drawn below. The vertex set is $V$ and the (symmetric) relation giving adjacency is $E$. Specifically, | + | Consider the graph $\mathcal{G} = (V,E)$ drawn below. The vertex set is $V$ and the (symmetric) relation giving adjacency is $E$. Specifically, |
- | Write down all elements of $\aut(\mathcal{G})$; | + | Write down all elements of $\aut(\mathcal{G})$; |
\begin{enumerate} | \begin{enumerate} | ||
- | \item There some $a\in \aut(\mathcal{G})$ such that $a^2$ is the identity. | + | \item There is an $a\in \aut(\mathcal{G})$ such that $a^2$ is the identity but $a$ is not the identity. |
- | \item There some $a\in \aut(\mathcal{G})$ such that $a^3$ is the identity. | + | \item There is an $a\in \aut(\mathcal{G})$ such that $a^3$ is the identity but $a$ is not the identity. |
- | \item There some $a\in \aut(\mathcal{G})$ such that every other $b\in \aut(\mathcal{G})$ is of the form $b=a^n$ for some $n$. | + | \item There is an $a\in \aut(\mathcal{G})$ such that every other $b\in \aut(\mathcal{G})$ is of the form $b=a^n$ for some $n$. |
\item For every $a,b\in \aut(\mathcal{G})$, | \item For every $a,b\in \aut(\mathcal{G})$, | ||
\end{enumerate} | \end{enumerate} | ||
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\end{center} | \end{center} | ||
\end{problem} | \end{problem} | ||
- | </code> | + | </file> |
---- | ---- | ||
==== External links ==== | ==== External links ==== | ||
* [[wp> | * [[wp> | ||
+ | |||
+ | {{tag> |