Math 120A HW 2 Solutions

1. The operation is not commutative because $c \ast a = c$, but $a \ast c = a$. The element $b$ is an identity element for the structure (this can be verified visually: the row and column corresponding to $b$ both list the elements $a$, $b$, $c$ in order.)
2. Yes, the function $f:S \to S$ defined by $f(0) = 1$ and $f(1) = 0$ is an isomorphism from $(S,\ast)$ to $(S,\ast')$. Clearly, $f$ is a bijection. To verify that $f$ is a homomorphism:
   • $f(0 \ast 0) = f(0) = 1$ and $f(0) \ast' f(0) = 1 \ast' 1 = 1$ also.
   • $f(0 \ast 1) = f(0) = 1$ and $f(0) \ast' f(1) = 1 \ast' 0 = 1$ also.
   • $f(1 \ast 0) = f(0) = 1$ and $f(1) \ast' f(0) = 0 \ast' 1 = 1$ also.
   • $f(1 \ast 1) = f(1) = 0$ and $f(1) \ast' f(1) = 0 \ast' 0 = 0$ also.
   You can also verify that $f$ is an isomorphism using the tables: take the table for $\ast$, replace every element (including those in row and column headers) with $f$ of that element, and now notice that the rows and columns can be reordered to give the table for $\ast'$.
3. For each of the following functions, say whether it is a homomorphism and also whether it is an isomorphism. Justify your answers.
   1. Here $\phi$ is a homomorphism because $\phi(m + n) = 3(m+n) = 3m+3n = \phi(m) + \phi(n)$ for all $m,n \in \mathbb{Z}$. It is not an isomorphism because it is not surjective; for example, the element $1$ is not in its range.
   2. Here $\phi$ is a homomorphism because $\phi(xy) = (xy)^3 = x^3y^3 = \phi(x)\phi(y)$ for all $x,y \in \mathbb{R}$. It is an isomorphism because it is also bijective (it is well-known that mapping $x$ to $x^n$ gives a bijection $\mathbb{R} \to \mathbb{R}$ if $n \in \mathbb{N}$ is odd: it is surjective because it is continuous and not bounded above or below, and it is injective because it is strictly increasing.)
   3. Here $\phi$ is not a homomorphism: $\phi(1+1) = 2^3 = 8$, but $\phi(1) + \phi(1) = 1^3 + 1^3 = 2$. So it is not an isomorphism.
4. One isomorphism from $(\mathbb{R}^*,\cdot)$ to $(S,\ast)$ is given by $$ \phi(a) = \begin{pmatrix} a & 0 \\ 0 & 1/a \end{pmatrix}. $$ This function $\phi$ is injective because if $a \ne b$ then the top-left entries of $\phi(a)$ and $\phi(b)$ are different. It is surjective because any matrix $A \in S$ is diagonal with diagonal entries $a$ and $b$, say, and $b = 1/a$ because the determinant is $1$, so $a = \phi(a)$. It is a homomorphism: for any $a,b \in \mathbb{R}^*$ we have $$\phi(ab) = \begin{pmatrix} ab & 0 \\ 0 & 1/ab \end{pmatrix} = \begin{pmatrix} ab & 0 \\ 0 & (1/a)(1/b) \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & 1/a \end{pmatrix} \begin{pmatrix} b & 0 \\ 0 & 1/b \end{pmatrix} = \phi(a)\phi(b).$$
5. Let $\phi$ be an isomorphism from a binary structure $(S,\ast)$ to a binary structure $(S',\ast')$. Assume that for every element $b\in S$ there is an element $a \in S$ such that $a \ast a = b$. We want to show that for every element $b' \in S'$ there is an element $a' \in S'$ such that $a' \ast' a' = b'$. Let $b' \in S'$. Take $b \in S$ such that $\phi(b) = b'$. Take $a \in S$ such that $a \ast a = b$. Define $a' = \phi(a)$. Then $a' \ast' a' = \phi(a) \ast' \phi(a) = \phi(a \ast a) = \phi(b) = b'$, as desired.
6. Using the structural property from the previous problem, prove the following statements:
   1. $(\mathbb{Z},+) \not\cong (\mathbb{Q}, +)$ because for every $b \in \mathbb{Q}$ there is an $a \in \mathbb{Q}$ such that $a + a = b$, namely $b/2$, but $1 \in \mathbb{Z}$ and there is no $a \in \mathbb{Z}$ such that $a + a = 1$.
   2. $(\mathbb{Q},+) \not\cong (\mathbb{Q},\cdot)$ because for every $b \in \mathbb{Q}$ there is an $a \in \mathbb{Q}$ such that $a + a = b$, namely $b/2$, but $2 \in \mathbb{Q}$ and there is no $a \in \mathbb{Q}$ such that $a \cdot a = 2$. (I hope you know a proof of this already, but here is one anyway: if not, write $a = m/n$ where $m$ and $n$ are relatively prime; if $a\cdot a = m^2/n^2 = 2$ then $m$ is even, so $m^2$ is divisible by $4$, which implies that $n$ is even also, a contradiction.)