Math 120A HW 4 Solutions
- 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 $a$.)
- 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 $n$.
- Set $G = \mathbb{Z}_{10}$ and $a = b = 1$. Then $a$ and $b$ have order $10$, but $a + b$ is $2$, which has order $5$.
- 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 $5$.
-
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 $3$.
- 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 $a$.
- Consider $S_4$, the symmetric group on the four-element set $\{1,2,3,4\}$.
- For example,
$$ \rho =
\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix}.
$$
- For example,
$$\sigma_1 =
\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 3 & 4 \end{pmatrix}
\quad\text{and}\quad
\sigma_2 =
\begin{pmatrix} 1 & 2 & 3 & 4\\ 1 & 2 & 4 & 3 \end{pmatrix}.$$
- For example,
$$\tau_1 =
\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 3 & 4 \end{pmatrix}
\quad\text{and}\quad
\tau_2 =
\begin{pmatrix} 1 & 2 & 3 & 4\\ 1 & 3 & 2 & 4 \end{pmatrix}.$$
- Calculate the orders of the following elements in the group $S_5$:
- The order of $\sigma$ is $4$.
- The order of $\tau$ is $6$.
-
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 $\sigma^{-1} \in H$.
- 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.)
-
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.)
-
We need to show that $D_\infty$ contains the identity permutation and is closed under inverses and multiplication.
- The identity permutation is included in $D_\infty$ as $\sigma_0$ (the "shift by $0$".)
- The inverse of $\sigma_n$ is $\sigma_{-n}$ (which is in $D_\infty$) because $(i+n) + (-n) = i$ and $(i+(-n)) + n = i$. The inverse of $\rho_n$ is itself because $n-(n-i) = i$ for all integers $i$.
- Now we show closure under multiplication. The product of two shifts $\sigma_m$ and $\sigma_n$ is the
shift $\sigma_{m+n} \in D_\infty$. Given two reflections $\rho_m$ and $\rho_n$ we
have $\rho_m(\rho_n(i)) = m - (n-i) = i + (m-n) = \sigma_{m-n}(i)$ for all integers $i$, so $\rho_m\rho_n =
\sigma_{m-n} \in D_\infty$. Finally, given a shift $\sigma_m$ and a reflection $\rho_n$ we have $\sigma_m(\rho_n(i)) = (n-i) + m = (n+m)-i = \rho_{n+m}(i)$ and
$\rho_n(\sigma_m(i)) = n-(i+m) = (n-m) - i = \rho_{n-m}(i)$ for all integers $i$, so $\sigma_m\rho_n = \rho_{n+m} \in D_\infty$ and $\rho_n\sigma_m = \rho_{n-m} \in D_\infty$.