Math 120A HW 7 Solutions

1. Let $x \in \{1,\ldots,n\}$. We want to show that $(\sigma_1\cdots \sigma_r)(x) = \sigma(x)$.
   • Case 1: $x$ is not an element of $B_i$ for any $i$. In this case $\sigma_i(x) = x$ for every $i$, and also $\sigma(x) = x$ (because if $\sigma(x) \ne x$ then the orbit of $x$ under $\sigma$ would have more than one element, and $x$ would be contained in some non-singleton orbit $B_i$.) So in this case we have $(\sigma_1\cdots \sigma_r)(x) = x = \sigma(x)$.
   • Case 2: $x \in B_i$ for some $i$. In this case, fixing such an $i$, we have $\sigma_i(x) = \sigma(x)$. Also in this case, $x$ is not an element of $B_j$ for any $j \ne i$, because the orbits of $\sigma$ are disjoint. Moreover, $\sigma(x)$ is not an element of $B_j$ for any $j \ne i$, because $\sigma(x)$ is in the same orbit as $x$. So we have $\sigma_j(x) = x$ and $\sigma_j(\sigma(x)) = \sigma(x)$ for all $j \ne i$. Therefore $(\sigma_1\cdots \sigma_r)(x) = (\sigma_1 \cdots \sigma_{i-1})(\sigma_i)(\sigma_{i+1}\cdots \sigma_r)(x) = (\sigma_1 \cdots \sigma_{i-1})(\sigma_i)(x) = (\sigma_1 \cdots \sigma_{i-1})(\sigma(x)) = \sigma(x)$.
2. First let's compute $\sigma = \begin{pmatrix} 1&2&3&4&5\\ 1&2&4&5&3 \end{pmatrix}$.
   1. $\sigma = (3\;4\;5)$.
   2. $\sigma = (3\;4)(4\;5)$ (for example.)
   3. No. In general, if a permutation $\rho$ is a product of disjoint transpositions $\tau_1\cdots\tau_n$, then $\rho^2 = (\tau_1\cdots\tau_n)^2 = \tau_1^2\cdots\tau_n^2 = \text{identity}$ because disjoint cycles commute and the square of every transposition is the identity. However, $\sigma^2 \ne \text{identity}$.
3. The elements of $A_4$ are the identity permutation, $(1\;2)(3\;4)$, $(1\;3)(2\;4)$, $(1\;4)(2\;3)$, $(2\;3\;4)$, $(2\;4\;3)$, $(1\;3\;4)$, $(1\;4\;3)$, $(1\;2\;4)$, $(1\;4\;2)$, $(1\;2\;3)$, and $(1\;3\;2)$.
4. In this problem we give geometric interpretations of the alternating groups $A_3$ and $A_4$.
   1. The elements of $A_3$ correspond to rotations by $0^\circ$, $120^\circ$, and $240^\circ$ around the center of the triangle. (Speaking in terms of algebra instead of geometry, you could say that the elements of $A_3$ are the rotations $\sigma_1$, $\sigma_2$, and $\sigma_3$ in $D_3$ as we have defined it.)
   2. No, for example the element $\sigma = (1\;2\;3) \in A_4$ does not correspond to any symmetry of the square, because the distance from vertex $1$ to vertex $4$ is not equal to the distance from vertex $\sigma(1)$ (${}= 2$) to vertex $\sigma(4)$ (${}= 4$.) Another way to see that the answer is "no" is to observe that $A_4$ has $12$ elements, but there are only $8$ symmetries of the square (equivalently, $D_4$ has only $8$ elements.)
   3. Every element of $A_4$ is either $(a\;b\;c)$ for some distinct $a,b,c \in \{1,2,3,4\}$, or $(a\;b)(c\;d)$ for some distinct $a,b,c,d \in \{1,2,3,4\}$, or the identity element. The symmetry corresponding to $(a\;b\;c)$ is a rotation by $120^\circ$ around the line through $d$ and the center of $\triangle abc$. (The rotation is in the direction from $a$ to $b$.) The symmetry corresponding to $(a\;b)(c\;d)$ is a rotation by $180^\circ$ around the line through the midpoints of $\overline{ab}$ and $\overline{cd}$. The symmetry corresponding to the identity element of $A_4$ is the identity function on Euclidean space, which you could describe as a rotation of $0^\circ$ around any axis.