Math 120A HW 8 Solutions

  1. Consider the group $G = \mathbb{Z}_9$ (under addition modulo $9$) and the cyclic subgroup $H = \langle 3 \rangle$. The group is abelian, so we can just talk about cosets rather than left or right cosets.
  2. Let $G$ be the group of all nonzero real numbers under multiplication and let $H$ be the cyclic subgroup of $G$ generated by $-2$. Let $a = 1$, $b = 1/2$, and $c = 1/3$. (Again, the group is abelian, so we can just talk about cosets rather than left or right cosets.)
    1. We have: $$\begin{align*} 1H &= \left\{ \ldots, -\frac{1}{8}, \frac{1}{4}, -\frac{1}{2}, 1, -2, 4, -8, \ldots \right\} \\ \frac{1}{2}H &= \left\{ \ldots, -\frac{1}{16}, \frac{1}{8}, -\frac{1}{4}, \frac{1}{2}, -1, 2, -4, \ldots \right\} \\ \frac{1}{3}H &= \left\{ \ldots, -\frac{1}{24}, \frac{1}{12}, -\frac{1}{6}, \frac{1}{3}, -\frac{2}{3}, \frac{4}{3}, -\frac{8}{3}, \ldots \right\} \\ \end{align*}$$
    2. We have $-4 \in \frac{1}{2}H$ (and of course $-4 \in -4H$,) so $-4H = \frac{1}{2}H$: the cosets of $H$ partition $G$, so if two cosets are not disjoint then they must be equal.

      Alternatively, you could prove directly that $-4H = \frac{1}{2}H$, using the observation that $-4(-2)^n = (-1)^{n+1}2^{n+2} = (-1)^{n+3}2^{n+2} = \frac{1}{2}(-2)^{n+3}$.

  3. Let $a,b,c \in G$ and assume that $a \sim_\text{L} b$ and $b \sim_\text{L} c$. In other words, $b \in aH$ and $c \in bH$. Take $h_1,h_2 \in H$ such that $b = ah_1$ and $c = bh_2$. Then $c = (ah_1)h_2 = a(h_1h_2) \in aH$ because $h_1h_2 \in H$, so $a \sim_\text{L} c$.
  4. Let $G$ be a group and let $H$ be a subgroup of $G$.
    1. Let $l$ be the number of left cosets of $H$ in $G$ and let $r$ be the number of right cosets of $H$ in $G$. Then we have $l|H| = |G| = r|H|$. Assume that $G$ is finite (so $H$, $l$, and $r$ are also finite.) Then we can divide by $|H|$ to get $l = |G|/|H| = r$.
    2. This argument doesn't work when $G$ and $H$ are infinite because it doesn't make sense to divide by the infinite number $|H|$. (Also, it is unjustified to cancel it from both sides of the equation $l|H| = r|H|$. It does make sense to multiply infinite numbers, by the way, but you don't need to learn that for this class.)
    3. We can define a function $f$ from the set of all left cosets of $H$ in $G$ to the set of all right cosets of $H$ in $G$ by $f(aH) = Ha^{-1}$. To see that there is such a function $f$ (i.e. $f$ is "well-defined") let $a,b\in G$ and assume that $aH = bH$; we want to show that $Ha^{-1} = Hb^{-1}$. Equivalently, assume that $a \in bH$; we want to show that $a^{-1} \in Hb^{-1}$. Take $h \in H$ such that $a = bh$. Then $a^{-1} = h^{-1}b^{-1} \in Hb^{-1}$ because $h^{-1} \in H$.

      (Alternatively, we can take inverses of both sides of the equation $aH = bH$, where by the inverse of a set $S$ we mean $\{x^{-1} : x \in S\}$, and then use the fact that $H^{-1} = H$.)

      The proof that $f$ is injective is similar. Let $a,b\in G$ and assume that $Ha^{-1} = Hb^{-1}$; we want to show that $aH = bH$. Equivalently, assume that $a^{-1} \in Hb^{-1}$; we want to show that $a \in bH$. Take $h \in H$ such that $a^{-1} = hb^{-1}$. Then $a = (a^{-1})^{-1} = (hb^{-1})^{-1} = (b^{-1})^{-1}h^{-1} = bh^{-1} \in bH$, because $h^{-1} \in H$.

      $f$ is surjective because for every $a \in G$ we have $f(a^{-1}H) = H(a^{-1})^{-1} = Ha$.

  5. Consider the group $\mathbb{Z}_3 \times \mathbb{Z}_3$ (with the operation of addition mod 3 in each coordinate.)
    1. We have:
      • $\langle (0,0) \rangle = \{ (0,0) \}$
      • $\langle (0,1) \rangle = \{ (0,0), (0,1), (0,2) \}$
      • $\langle (0,2) \rangle = \{ (0,0), (0,2), (0,1) \}$ (${} = \langle (0,1) \rangle$)
      • $\langle (1,0) \rangle = \{ (0,0), (1,0), (2,0) \}$
      • $\langle (1,1) \rangle = \{ (0,0), (1,1), (2,2) \}$
      • $\langle (1,2) \rangle = \{ (0,0), (1,2), (2,1) \}$
      • $\langle (2,0) \rangle = \{ (0,0), (2,0), (1,0) \}$ (${} = \langle (1,0) \rangle$)
      • $\langle (2,1) \rangle = \{ (0,0), (2,1), (1,2) \}$ (${} = \langle (1,2) \rangle$)
      • $\langle (2,2) \rangle = \{ (0,0), (2,2), (1,1) \}$ (${} = \langle (1,1) \rangle$)
    2. No. We could simply observe that the order of $(a,b)$ in $\mathbb{Z}_3 \times \mathbb{Z}_3$ is the least common multiple of the order of $a$ in $\mathbb{Z}_3$ and the order of $b$ in $\mathbb{Z}_3$, which cannot be equal to $9$ (it must be $1$ or $3$; see the more general argument in the next part.)
    3. The order of $(a,b)$ in $\mathbb{Z}_p \times \mathbb{Z}_p$ is the least common multiple of the order of $a$ in $\mathbb{Z}_p$ and the order of $b$ in $\mathbb{Z}_p$. Each of these orders divides $p$ by Lagrange's theorem applied to the group $\mathbb{Z}_p$. So each of these orders must be $1$ or $p$ because $p$ is prime, and therefore their least common multiple is $1$ or $p$.

      (It is $1$ if $a$ and $b$ are both zero; otherwise it is $p$.)

  6. Assume toward a contradiction that $\mathbb{Z} \times \mathbb{Z}_2$ is cyclic. Then it must be isomorphic to $\mathbb{Z}$. (This is because every cyclic group is isomorphic to $\mathbb{Z}$ or to $\mathbb{Z}_n$ for some positive integer $n$, according to whether it has infinite order or finite order $n$, and clearly $\mathbb{Z} \times \mathbb{Z}_2$ has infinite order.) Note that the element $(0,1)$ of $\mathbb{Z} \times \mathbb{Z}_2$ has order $2$. Therefore the isomorphic group $\mathbb{Z}$ must also have an element of order $2$. But this contradicts the fact that every element of $\mathbb{Z}$ has order $1$ or infinite order.

    Alternatively, letting $(a,b) \in \mathbb{Z} \times \mathbb{Z}_2$, we can show directly that $\langle (a,b) \rangle \ne \mathbb{Z} \times \mathbb{Z}_2$: in the case $a = 0$ we have $(1,0) \notin \langle (a,b) \rangle$, and in the case $a \ne 0$ we have $(0,1) \notin \langle (a,b) \rangle$. (For the second case $a \ne 0$, note that if $n(a,b) = (0,1)$ then $na = 0$, so $n = 0$, so $nb = 0$, so $0 = 1$, a contradiction.)