Compute the following compositions:
\(\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}\)
\(\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1 \end{pmatrix}\)
\(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 2 & 1 & 6 & 4 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 2 & 1 & 4 & 3 & 6 & 4 \end{pmatrix}\)
Solution:
We compute that \[\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix}\] and \[\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4\\ 3 & 4 & 1 & 2 \end{pmatrix}\] and \[\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 2 & 1 & 6 & 4 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 2 & 1 & 4 & 3 & 6 & 4 \end{pmatrix}= \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 5 &3 & 1 & 2 & 4 & 1 \end{pmatrix}.\]
Compute the inverses of the following permutations:
\(\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}\)
\(\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3 \end{pmatrix}\)
\(\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 2 & 1 & 6 & 4 \end{pmatrix}\)
Solution:
As illustrated in the solution to Problem 1, we see that the inverse of (a) is itself. The inverse of (b) is also itself. And the inverse of (c) is \[\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 4 & 3 & 1 & 6 & 2 & 5 \end{pmatrix}.\]
Let \(P_n\) be a regular polygon with \(n\) sides where \(n\geq 4\). Convince yourself that the symmetry group of \(P_n\) is not the same as the symmetric group on \(\{1,\dots,n\}\).
Hint: how many elements are in \(\text{Sym}(P_n)\) and \(S_n\)?
Solution:
The argument for Exercise 4 also holds here. In other words, one reason to see why the symmetry group of \(P_n\) is not the same as the symmetric group on \(\{1,\dots,n\}\) is to realize that the number of elements in \(\text{Sym}(P_n)\) differs from the number of elements in \(S_n\). In fact, there are \(2n\) elements in \(\text{Sym}(P_n)\) and \(n!\) elements in \(S_n\). And we see that \(2n < n!\) when \(n\geq 4\).