Math 120A HW 6 Solutions

  1. Assume that $a \in H+b$. This means that $a = h+b$ for some $h \in H$. So given an element of $H+a$, say $h' + a$ where $h' \in H$, we have $h' + a = h' + h + b \in H+b$ because $h' + h \in H$. Conversely, given an element of $H + b$, say $h' + b$ where $h' \in H$, noting that $(-h) + a = b$, we have we have $h' + b = h' - h + a \in H+a$ because $h'-h \in H$. Therefore the cosets $H + a$ and $H+b$ have the same elements, so they are equal.
  2. The cosets of $15\mathbb{Z}$ in $5\mathbb{Z}$ are $15\mathbb{Z}$, $15\mathbb{Z} + 5$, and $15\mathbb{Z} + 10$, so there are three cosets. (To see that this is all of them, note that if $a \in 5 \mathbb{Z}$ then the coset $15\mathbb{Z} + a$ is equal to one of these three cosets listed, according to whether $a \mathbin{\text{mod}} 15$ is $0$, $5$, or $10$.)
  3. Let $G$ be a group, let $H$ be a subgroup of $G$, and let $K$ be a subgroup of $H$ (so $K$ is also a subgroup of $G$.)
    1. If $G$ is finite, then we have \[ (G : H)(H : K) = \frac{|G|}{|H|}\cdot \frac{|H|}{|K|} = \frac{|G|}{|K|} = (G : K).\] Note: the equation $(G : H) = |G|/|H|$ is NOT the defintion of the index $(G : H)$, it is just a useful equation that happens to be true if $G$ is finite. If $G$ is infinite then $|G|/|H|$ is meaningless.
    2. We have $(\mathbb{Z} : 5\mathbb{Z})(5\mathbb{Z} : 15\mathbb{Z}) = (\mathbb{Z} : 15\mathbb{Z})$, and because the number of cosets of $n \mathbb{Z}$ in $\mathbb{Z}$ is $n$ we have $(\mathbb{Z} : 5\mathbb{Z}) = 5$ and $(\mathbb{Z} : 15\mathbb{Z}) = 15$, so $5(5\mathbb{Z} : 15\mathbb{Z}) = 15$, and therefore $(5\mathbb{Z} : 15\mathbb{Z}) =3$.
  4. Consider the cyclic subgroup $H$ of $S_3$ generated by the cycle $(1\;2\;3)$.
    1. The index of $H$ in $S_3$ is given by $(S_3 : H) = |S_3|/|H| = 6/3 = 2$.
    2. The left cosets of $H$ in $S_3$ are given by \begin{align*} H &= \{\text{identity}, (1\;2\;3), (1\;3\;2)\}\quad\text{and}\\ (1\;2)H &= \{(1\; 2), (2\;3), (1,3)\}. \end{align*} (Note that there is nothing special about using $(1\;2)$ here, but if we had tried an element of $H$ instead we just would have gotten $H$ again.)
    3. The right cosets of $H$ in $S_3$ are given by \begin{align*} H &= \{\text{identity}, (1\;2\;3), (1\;3\;2)\}\quad\text{and}\\ H(1\;2) &= \{(1\; 2), (1\;3), (2,3)\}. \end{align*} (Note that the elements of our second coset appear in a different order than in the previous part of the problem because $(1\;2)$ does not commute with all the elements of $H$. But this does not matter; the cosets themselves are the same.)
    4. Assume that $H$ has index $2$ in $G$ and let $a \in G$. If $a \in H$ then $aH = H = Ha$. If $a \notin H$ then $aH$ is not equal to $H$, and because the left cosets partition $G$ into two parts, $aH$ must be the complement of $H$ in $G$. A similar argument shows that $Ha$ is also the complement of $H$ in $G$, so again $aH = Ha$ as desired.
  5. First, we have $\mathbb{Z}_2 \times \mathbb{Z}_4$ itself and the trivial subgroup $\{(0,0)\}$. By Lagrange's theorem the only other possible orders of subgroups are 2 and 4. Every subgroup of order 2 is cyclic, so we just need to find the elements of $\mathbb{Z}_2 \times \mathbb{Z}_4$ of order $2$ and look at the cyclic subgroups they generate. This gives us the subgroups $\langle (1,0)\rangle$, $\langle (0,2)\rangle$, and $\langle (1,2)\rangle$. For the cyclic subgroups of order 4, we just need to find the elements of $\mathbb{Z}_2 \times \mathbb{Z}_4$ of order $4$ and look at the cyclic subgroups they generate. This gives us the subgroups $\langle (0,1)\rangle$ and $\langle (1,1)\rangle$. (There are two other elements of order $4$ but they generate the same two subgroups.) Finally for the non-cyclic subgroups of order 4, their three non-identity elements must have order 2, so the only possibility is $\{(0,0),(0,2),(1,0),(1,2)\} = \{0,1\} \times \{0,2\}$. These eight subgroups that we found are shown in a diagram. (Remark: $\{0,1\} \times \{0,2\}$ is isomorphic to the Klein group $V$, so below it in the diagram we can see a copy of the subgroup diagram for $V$.)
  6. The group $\mathbb{Z}_3 \times \mathbb{Z}$ has an element of order 3, for example $(1,0)$. So if it were isomorphic to the group $\mathbb{Z}_2 \times \mathbb{Z}$, the latter group would also have an element of order 3, for example $\phi((1,0))$ where $\phi$ is such an isomorphism. But the elements of $\mathbb{Z}_2 \times \mathbb{Z}$ either have order 1 or 2 (if their second coordinate is zero) or infinite order (if their second coordinate is nonzero.) So there can be no such isomorphism. (Note that the groups themselves have the same order, meaning that there is a bijection between them, so a cardinality argument won't work.)