Grade 9/10 Math Circles
An Introduction to Group Theory Part 2 - Problem Set

1. Compute the following compositions:

   1. $\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)\circ \left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$

   2. $\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 1& 4& 3\end{array}\right)\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 3& 2& 1\end{array}\right)$

   3. $\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 5& 2& 1& 6& 4\end{array}\right)\circ \left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 2& 1& 4& 3& 6& 4\end{array}\right)$

   Solution:

   We compute that $$\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)\circ \left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)=\left(\begin{array}{cc}1& 2\\ 1& 2\end{array}\right)$$ and $$\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 1& 4& 3\end{array}\right)\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 3& 2& 1\end{array}\right)=\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 4& 1& 2\end{array}\right)$$ and $$\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 5& 2& 1& 6& 4\end{array}\right)\circ \left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 2& 1& 4& 3& 6& 4\end{array}\right)=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 5& 3& 1& 2& 4& 1\end{array}\right).$$

2. Compute the inverses of the following permutations:

   1. $\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$

   2. $\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 1& 4& 3\end{array}\right)$

   3. $\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 5& 2& 1& 6& 4\end{array}\right)$

   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 $$\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 4& 3& 1& 6& 2& 5\end{array}\right).$$

3. Let $P_n$ be a regular polygon with $n$ sides where $n\ge 4$. 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 $n\ge 4$.