Math 120A HW 4 Solutions

1. Recall that $\mathbb{Z}_n$ denotes the group $(\mathbb{Z}_n,+_n)$.
   1. $\mathbb{Z}_3$ has no nontrivial proper subgroup. If $H$ is a nontrivial subgroup of $\mathbb{Z}_3$ then it contains the identity element and some other element, either $1$ or $2$. If it contains $1$ then it contains $1 +_3 1$, which is $2$, and if it contains $2$ then it contains $2 +_3 2$, which is $1$, and in either case it is improper.
   2. $\mathbb{Z}_4$ has a nontrivial proper subgroup $\{0,2\}$. Because $0$ is the identity element, to check that this is a subgroup we only need to check that $2 +_4 2 = 0$.
2. In this problem we consider intersections and unions of subgroups.
   1. To verify closure under the operation, let $a,b \in H \cap K$. Then $a \ast b \in H$ because $H$ is closed under $\ast$, and $a \ast b \in K$ because $K$ is closed under $\ast$, so $a \ast b \in H \cap K$. Because the identity element of $G$ is in $H$ and also in $K$, it is in $H \cap K$. To verify closure under inverses, let $a \in H \cap K$. Then $a^{-1} \in H$ because $H$ is closed under inverses, and $a^{-1} \in K$ because $K$ is closed under inverses, so $a^{-1} \in H \cap K$.
   2. Define $H = \{0,2,4\}$ and $K = \{0,3\}$ (it's easy to check that these are subgroups of $\mathbb{Z}_6$.) Then $H \cup K = \{0,2,3,4\}$, which is not a subgroup because it contains $2$ and $3$ but not $2 +_6 3$, which is $5$.
3. For this problem only, you do not need to justify your answers.
   1. The subgroups of $\mathbb{Z}_4$ are $\{0\}$, $\{0,2\}$, and $\mathbb{Z}_4$ itself.
   2. The subgroups of $\mathbb{Z}_5$ are $\{0\}$ and $\mathbb{Z}_5$.
   3. The subgroups of $\mathbb{Z}_6$ are $\{0\}$, $\{0,2,4\}$, $\{0,3\}$, and $\mathbb{Z}_6$.
   (It's hard for me to make diagrams here; for examples of subgroup diagrams, see the exam solutions.)
4. Let $G$ and $G'$ be groups, let $\phi$ be a homomorphism from $G$ to $G'$, and let $a \in G$. Because we haven't given names to the operations of $G$ and $G'$, we default to using multiplicative notation.
   1. Let $n \in \mathbb{Z}$. We split into cases according to the definition of $a^n$. If $n \gt 0$ then \[\phi(a^n) = \phi(\underbrace{a \cdots a}_{n\text{ times}}) = \underbrace{\phi(a)\cdots \phi(a)}_{n\text{ times}} = \phi(a)^n.\] (A more formal proof would use induction on $n$.) If $n = 0$ then $\phi(a^n) = \phi(e_G) = e_{G'} = \phi(a)^n$. If $n \lt 0$ then $\phi(a^n) = \phi((a^{-1})^{-n}) = \phi(a^{-1})^{-n} = (\phi(a)^{-1})^{-n} = \phi(a)^n$ (this uses what we have already proved for positive numbers and in particular for $-n$.)
   2. $\phi[\langle a \rangle] = \phi[ \{ a^n : n \in \mathbb{Z} \}] = \{\phi(a^n) : n \in \mathbb{Z}\} = \{\phi(a)^n : n \in \mathbb{Z}\} = \langle \phi(a) \rangle$.
   3. $\log[\langle a \rangle] = \log[ \{ a^n : n \in \mathbb{Z} \}] = \{\log(a^n) : n \in \mathbb{Z}\} = \{n \log(a) : n \in \mathbb{Z}\} = \langle \log(a) \rangle$.
5. To understand a group it is helpful to know the orders of its elements, as we will see in this problem.
   1. We compute successive multiples in the sense of the operation $+_{10}$ (not called powers, because we are using additive notation) until we get the identity element, which is $0$:
      • The order of $0$ is $1$ because it is the identity element.
      • Successive multiples of $1$ are $1,2,3,4,5,6,7,8,9,0$, so the order of $1$ is $10$.
      • Successive multiples of $2$ are $2,4,6,8,0$, so the order of $2$ is $5$.
      • Successive multiples of $3$ are $3,6,9,2,5,8,1,4,7,0$, so the order of $3$ is $10$.
      • Successive multiples of $4$ are $4,8,2,6,0$, so the order of $4$ is $5$.
      • Successive multiples of $5$ are $5,0$, so the order of $5$ is $2$.
      • Successive multiples of $6$ are $6,2,8,4,0$, so the order of $6$ is $5$.
      • Successive multiples of $7$ are $7,4,1,8,5,2,9,6,3,0$, so the order of $7$ is $10$.
      • Successive multiples of $8$ are $8,6,4,2,0$, so the order of $8$ is $5$.
      • Successive multiples of $9$ are $9,8,7,6,5,4,3,2,1,0$, so the order of $9$ is $10$.
   2. Let $H$ be a subgroup of $\mathbb{Z}_{10}$. If $H$ were isomorphic to $\mathbb{Z}_3$ then $H$ would have an element of order $3$, because $\mathbb{Z}_3$ has an element of order $3$ and having an element of order $3$ is a structural property. (More specifically, if $\phi$ were an isomorphism from $\mathbb{Z}_3$ to $H$ then $\phi(1)$ would have order $3$ in $H$.) However, we have shown that $\mathbb{Z}_{10}$ has no element of order $3$. (Note that the order of an element of the subgroup $H$ in the sense of $H$ is the same as the order of that element in the sense of the larger group $\mathbb{Z}_{10}$.)

      Another way we can do this problem now is to notice that because $\mathbb{Z}_{10}$ is cyclic, all of its subgroups are cyclic. We computed all of its cyclic subgroups in part (a), and none of them has order $3$, so none of them can be isomorphic to $\mathbb{Z}_3$.

6. Let $A$ be a $2 \times 2$ diagonal matrix with diagonal entries $a_{1,1},a_{2,2} \gt 0$. We want to show that $A$ does not generate $G$. (There are many ways to do this. Here is one.) If $a_{1,1}$ and $a_{2,2}$ are both equal to $1$ then clearly $A$ does not generate $G$, so without loss of generality we may assume that $a_{1,1} \ne 1$ (the proof in the case $a_{2,2} \ne 1$ is similar.) Note that $A^n$ is the $2 \times 2$ diagonal matrix with diagonal entries $(a_{1,1})^n$ and $(a_{2,2})^n$. Define $B$ to be the $2 \times 2$ diagonal matrix with diagonal entries $b_{1,1} = a_{1,1}$ and $b_{2,2} = a_{2,2} + 1$ (any positive number other than $a_{2,2}$ would work here.) Then $B \in G$, but $B \ne A^n$ for all $n \in \mathbb{Z}$: if $A^n = B$ then $(a_{1,1})^n = b_{1,1} = a_{1,1}$, which implies $n = 1$ because $a_{1,1}$ is positive and not equal to $1$, so $A = B$, which clearly contradicts the definition of $B$. Therefore $A$ does not generate $G$.