Math 120A HW 3 Solutions

1. An automorphism of a group is an isomorphism of that group with itself.
   1. There are two automorphisms of $\mathbb{Z}_4$: the identity function and the function $f$ defined by $f(x) = -x \mathbin{\text{mod}} 4$. (To see that this is all of them, note that an isomorphism must map a generator to a generator. Therefore an automorphism of $\mathbb{Z}_4$ must map $1$ to either $1$ or $3$, and either of these choices completely determines the other values of the automorphism.)
   2. If we define a function $f: \mathbb{Z}_4 \to \mathbb{Z}_4$ by $f(0) = 0$, $f(1) = 2$, $f(2) = 1$, and $f(3) = 3$, this is a bijection but it is not an isomorphism because $f(1 +_4 1) = 1$ but $f(1) +_4 f(1) = 0$.
2. Assume that $a^5 = e$ and $a^3 = e$. Then $a^6 = (a^3)^2 = e^2 = e$, and $a^{-5} = (a^5)^{-1} = e^{-1} = e$, so we have $a = a^6a^{-5} = ee = e$.
3. The four conditions to verify are associativity, commutativity, existence of an identity element, and existence of inverses.
   • For associativity, let $A,B,C \in \mathcal{P}(U)$. The sets $(A \mathbin{\triangle} B) \mathbin{\triangle} C$ and $A \mathbin{\triangle} (B \mathbin{\triangle} C)$ are equal because they have the same elements: Letting $x \in U$, if $x$ is in $1$ or $3$ of the sets $A$, $B$, and $C$ then $x \in (A \mathbin{\triangle} B) \mathbin{\triangle} C$ and $x \in A \mathbin{\triangle} (B \mathbin{\triangle} C)$, and if $x$ is in $0$ or $2$ of the sets $A$, $B$, and $C$ then $x \notin (A \mathbin{\triangle} B) \mathbin{\triangle} C$ and $x \notin A \mathbin{\triangle} (B \mathbin{\triangle} C)$. (All eight subcases can be verified by inspection, so I don't expect you to write any more than this.)
   • For commutativity, let $A,B \in \mathcal{P}(U)$ and note that $A \mathbin{\triangle} B = (A \cup B) - (A \cap B) = (B \cup A) - (B \cap A) = B \mathbin{\triangle} A$.
   • The empty set is an identity element: For $A \in \mathcal{P}(U)$ we have $A \mathbin{\triangle} \emptyset = (A \cup \emptyset) - (A \cap \emptyset) = A - \emptyset = A$. (This is sufficient by commutativity.)
   • Every element $A \in \mathcal{P}(U)$ has an inverse, namely $A$ itself: $A \mathbin{\triangle} A = (A \cup A) - (A \cap A) = A - A = \emptyset$, which is the identity. (This is sufficient by commutativity.)
4. Besides $\mathbb{Z}_{10}$ itself and its trivial subgroup $\{0\}$, we have the subgroups $\{0,2,4,6,8\}$ and $\{0,5\}$. These are represented in a diagram with $\mathbb{Z}_{10}$ at the top, with $\{0,2,4,6,8\}$ and $\{0,5\}$ side-by-side below it, and $\{0\}$ at the bottom. There are lines from $\mathbb{Z}_{10}$ to $\{0,2,4,6,8\}$ and $\{0,5\}$, and there are lines from $\{0,2,4,6,8\}$ and $\{0,5\}$ to $\{0\}$.
5. Suppose that $\langle a \rangle \subseteq \langle b \rangle$. Then we have $a \in \langle a \rangle \subseteq \langle b \rangle$, so $a = nb$ for some $n \in \mathbb{Z}$, which means that $b \mid a$. For the converse, suppose that $b \mid a$. Fix $n \in \mathbb{Z}$ such that $a = nb$. For every $c \in \langle a \rangle$ we have $c = ma$ for some $n \in \mathbb{Z}$, so $c = m(nb) = (mn)b \in \langle b \rangle$. Therefore $\langle a \rangle \subseteq \langle b \rangle$.
6. A group $G$ is defined to be cyclic if there is an element $a \in G$ such that $\langle a \rangle = G$.
   1. $(\mathbb{Q}, +)$ is not cyclic: Let $a \in \mathbb{Q}$. If $a = 0$ then $\langle a \rangle = \{0\} \ne \mathbb{Q}$, and if $a \ne 0$ then $a/2 \notin \langle a \rangle$, so again $\langle a \rangle \ne \mathbb{Q}$.
   2. $(\mathbb{Z} \times \mathbb{Z}, +)$ is not cyclic: Consider $(a,b) \in \mathbb{Z} \times \mathbb{Z}$. If $a = 0$ then $(1,0) \notin \langle (a,b) \rangle$. If $b = 0$ then $(0,1) \notin \langle (a,b) \rangle$. If $a$ and $b$ have opposite signs then $(1,1) \notin \langle (a,b) \rangle$. If $a$ and $b$ have the same sign then $(1,-1) \notin \langle (a,b) \rangle$. So in every case we have $\langle (a,b) \rangle \ne \mathbb{Z} \times \mathbb{Z}$.
7. $g^{-1} = (ab^2a^3ba^4)^{-1} = a^{-4}b^{-1}a^{-3}b^{-2}a^{-1} = ab^2a^2ba^4$.
8. Let $i,j \in \mathbb{Z}$ and assume that $i \ne j$. Without loss of generality, we may assume that $i \lt j$. Then $j-i$ is a positive integer, so by our hypothesis we have $a^{j-i} \ne e$, and multiplying both sides by $a^i$ gives $a^j \ne a^i$.
9. Successive multiples of $5$ mod $10$ are $5$ and $0$, so $5$ has order $2$. Successive multiples of $6$ mod $10$ are $6,2,8,4,0$, so $6$ has order $5$. Successive multiples of $7$ mod $10$ are $7,4,1,8,5,2,9,6,3,0$, so $7$ has order $10$.