Math 120A HW 4 Solutions

  1. Let $(G, +)$ be an abelian group and let $a, b \in G$. (Here we follow the "additive notation" convention of denoting the operation of an arbitrary abelian group by "$+$". Accordingly, we write $0$ for the identity and write $na$ instead of $a^n$, for example in the definition of the order of $a$.)
    1. Assume that $a$ and $b$ have order $n$. Then $n(a+b) = na + nb = 0 + 0 = 0$, so the order of $a+b$ is less than or equal to $n$.
    2. Set $G = \mathbb{Z}_{10}$ and $a = b = 1$. Then $a$ and $b$ have order $10$, but $a + b$ is $2$, which has order $5$.
    3. Set $G = \mathbb{Z}_{10}$ and $a = b = 6$. Then $a$ and $b$ have order $5$, and $a+b$ is $2$, which also has order $5$.
  2. Define $f$ to be the permutation switching $x$ and $y$ and define $g$ to be the permutation switching $y$ and $z$. (Clearly both $f$ and $g$ have order $2$.) Then we have $(g \circ f)(x) = g(f(x)) = g(y) = z$, $(g \circ f)(y) = g(f(y)) = g(x) = x$, and $(g \circ f)(z) = g(f(z)) = g(z) = y$, so it's easy to see that the permutation $g \circ f$ has order $3$.
  3. Let $H$ be a subgroup of $\mathbb{Z}$. If $H$ is trivial then it is generated by $0$ and we're done. If not, it has a nonzero element. Because it is closed under negation, it has a positive element. By the well-ordering property of $\mathbb{Z}^+$, it has a least positive element $a$. We claim that $H = \langle a \rangle$. Because $H$ is a subgroup containing $a$, we have $\langle a \rangle \subseteq H$. Assume toward a contradiction that $H$ has an element $b \notin \langle a \rangle$. Then $na \lt b \lt n(a+1)$ for some $n \in \mathbb{Z}$. Subtracting $na$, we get $0 \lt b-na \lt a$. Then $b - na$ is a positive element of $H$ less than $a$, contradicting the definition of $a$.
  4. Consider $S_4$, the symmetric group on the four-element set $\{1,2,3,4\}$.
    1. For example, $$ \rho = \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix}. $$
    2. For example, $$\sigma_1 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 3 & 4 \end{pmatrix} \quad\text{and}\quad \sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 1 & 2 & 4 & 3 \end{pmatrix}.$$
    3. For example, $$\tau_1 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 3 & 4 \end{pmatrix} \quad\text{and}\quad \tau_2 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 1 & 3 & 2 & 4 \end{pmatrix}.$$
  5. Calculate the orders of the following elements in the group $S_5$:
    1. The order of $\sigma$ is $4$.
    2. The order of $\tau$ is $6$.
  6. For $\sigma, \tau \in H$ we have $(\sigma \circ \tau)(x) = \sigma(\tau(x)) = \sigma(x) = x$, so $\sigma \circ \tau \in H$. Clearly the identity permutation is in $H$. For $\sigma \in H$ we have $\sigma(x) = x$, and applying $\sigma^{-1}$ to both sides we get $x = \sigma^{-1}(x)$, so $\sigma^{-1} \in H$.
  7. Two different reflections in $D_4$ commute with one another if and only if their lines of reflection are at right angles to one another. (In fact, this holds for reflections in general.)
  8. For example, the reflections $\rho_1$ and $\rho_2$ across the lines through the vertices $1$ and $2$ respectively are given by $$\rho_1 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5\\ 1 & 5 & 4 & 3 & 2\end{pmatrix} \quad\text{and}\quad \rho_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5\\ 3 & 2 & 1 & 5 & 4\end{pmatrix}.$$ Their product $\rho_1\rho_2$ is given by $$\rho_1\rho_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5\\ 4 & 5 & 1 & 2 & 3\end{pmatrix}.$$ It is a rotation by $2/5$ turn, i.e. $144^\circ$. (More generally, the angle of rotation will be twice the angle between the two lines of reflection.)
  9. We need to show that $D_\infty$ contains the identity permutation and is closed under inverses and multiplication.