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Problem 7 (Automorphisms Of A Square 2)

Again consider the graph $\mathcal{G} = (V,E)$ shown below with vertex set $V = \{1,2,3,4\}$ and edges $$E = \{\{1,2\},\{2,3\},\{3,4\},\{1,4\}\}.$$

Recall that set of all automorphisms of $\mathcal{G}$ is written $\aut(\mathcal{G})$. We listed all the elements in this set in problem AutomorphismsOfASquare. Determine if each statement below is true or false. Make sure you justify your answers.

  1. There is an automorphism $a\in \aut(\mathcal{G})$ such that $a^2=a\circ a$ (the composition) is the identity automorphism, but $a$ is not the identity.
  2. There is an $a\in \aut(\mathcal{G})$ such that $a^3=a\circ a\circ a$ is the identity but $a$ is not the identity.
  3. 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 $n$. [Hint: for each element in $\aut(G)$, find the smallest $n$ such that $a^n=id_V$. ]
  4. For every $a,b\in \aut(\mathcal{G})$, we have $a\circ b = b\circ a$.

Solution

We first name each automorphism of the graph. The automorphisms of the graph are $$\begin{align} a_o&=\begin{pmatrix}1&2&3&4\\1&2&3&4\end{pmatrix},&&& a_1&=\begin{pmatrix}1&2&3&4\\2&3&4&1\end{pmatrix},\\ a_2&=\begin{pmatrix}1&2&3&4\\3&4&1&2\end{pmatrix},&&& a_3&=\begin{pmatrix}1&2&3&4\\4&1&2&3\end{pmatrix}, \\ a_4&=\begin{pmatrix}1&2&3&4\\3&2&1&4\end{pmatrix},&&& a_5&=\begin{pmatrix}1&2&3&4\\4&3&2&1\end{pmatrix}, \\ a_6&=\begin{pmatrix}1&2&3&4\\1&4&3&2\end{pmatrix},&&& a_7&=\begin{pmatrix}1&2&3&4\\2&1&4&3\end{pmatrix}. \end{align}$$

We now consider the statements below.

1. I claim there exists an automorphism $a\in\aut(\mathcal{G})$ such that $a^2=a\circ a$ is the identity automorphism, but $a$ is not the identity. The automorphism $a_2$ satisfies this, as it is not the identity automorphism but $a_2\circ a_2$ is the identity automorphism.

2. I claim There does not exist an $a\in\aut(\mathcal{G})$ such that $a^3=a\circ a\circ a$ is the identity automorphism but $a$ is not the identity. We will look at each automorphism. The automorphism $a_o$ cannot satisfy the statement because $a_o$ is the identity. We have that $$\begin{align} a_1^3&=\begin{pmatrix}1&2&3&4\\4&1&2&3\end{pmatrix},&&& a_2^3&=\begin{pmatrix}1&2&3&4\\3&4&1&2\end{pmatrix}, \\ a_3^3&=\begin{pmatrix}1&2&3&4\\2&3&4&1\end{pmatrix},&&& a_4^3&=\begin{pmatrix}1&2&3&4\\3&2&1&4\end{pmatrix}, \\ a_5^3&=\begin{pmatrix}1&2&3&4\\4&3&2&1\end{pmatrix},&&& a_6^3&=\begin{pmatrix}1&2&3&4\\1&4&3&2\end{pmatrix}, \\ a_7^3&=\begin{pmatrix}1&2&3&4\\2&1&4&3\end{pmatrix}. \end{align}$$ Thus, there is not an $a\in\aut(\mathcal{G})$ such that $a^3$ is the identity automorphism but $a$ is not the identity.

3. I claim There is not an $a\in\aut(\mathcal{G})$ such that every other $b\in\aut(\mathcal{G})$ can be written in the form $b=a^n$ for some $n$. Recall there are a total of eight automorphisms on this graph. We will show that all the automorphisms are of order less than eight, which means that none of the automporphisms can be used to acquire all eight. First, we know $a_o$ is the identity automorphism and is of order 1. Next, we know the automorphisms $a_2,a_4,a_5,a_6$, and $a_7$ are of order 2. Lastly, we know the automorphisms $a_1$ and $a_3$ are of order 4. Thus, because all automorphisms are of order less than eight, there are no automorphisms for which every other automorphism can be written as a power of that automorphism.

4. I claim that we do not have $a\circ b=b\circ a$ for every $a,b\in\aut(\mathcal{G})$. Let $a=a_1$ and $b=a_4$. Thus we have \begin{equation} a\circ b = a_1\circ a_4 = \begin{pmatrix}1&2&3&4\\2&1&4&3\end{pmatrix} \neq \begin{pmatrix}1&2&3&4\\4&3&2&1\end{pmatrix} = a_4\circ a_1 = b\circ a. \end{equation}

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